6:15 In this Case Let initial velocity (u) is 20km/hr=5.55m/s Angle is 30° Gravity(g) is 9.8m/s^2 Range that is horizontal distance covered is, R=u^2xsin2(30°)/g =5.55x5.55x0.866/9.8 =2.72 Matlab thele par se gaadi sirf 2.72 meter aage jake giregi. Height attained by scooty is, H=u^2xsin^2(30°)/2g =5.55x5.55x0.25/19.6 =0.4 meter Matlab scooty Gaadi ko jake thokni chahiye. Time Required is, T=2usin(30°)/g =2x5.55x0.5/9.8 =0.6 seconds Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9 seconds💀💀 (1 like to banata hai) Edit: Nishchay Bhai please ♥️
6:41 Given: - Mass of the object (m) = 350 kg - Initial speed (u) = 20 km/h - Angle of projection (θ) = 30° First, we need to convert the initial speed from km/h to m/s: 1 km/h = 1000 m / 3600 s = 5/18 m/s So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s Now, we can calculate the horizontal and vertical components of the object's velocity: Horizontal component (ux) = u * cos(θ) Vertical component (uy) = u * sin(θ) ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s Next, we can calculate the time of flight (t) using the vertical motion: The time to reach the highest point (time to peak) is given by: t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity) t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s Since the motion is symmetrical, the total time of flight is twice the time to reach the peak: t = 2 * 0.283 s = 0.566 s Finally, we can find the horizontal distance (range) the object will travel: Range = ux * t Range = 4.81 m/s * 0.566 s ≈ 2.72 meters Therefore, the object will land approximately 2.72 meters away from the point where it was launched. Answer: The object will land approximately 2.72 meters away from the launch point.
6:37 Velocity = 20 km/hr = 20×1000/3600 = 50/9 m/s = 5.55 m/s Angle of elevation with the ground= 30° Range= (5.55)²sin(2×30°)/(9.8) ≈ 2.72201 metres (Its the maximum range, g is taken constant, and the mass is neglected. If mass is also taken into account then the range would be less than this)
6:37 at this point you asked us a ques Ans. u=20km/h=5.5m/s g=10m/s*2 Range of a projectile= u^2*sin2theta/g =5.5^2 x sin60/10 =3.025 x √3/2 =2.6165 Pls like, it took some effort to calculate ❤❤❤❤ Length of 1 car= 4.8m Length of 5 car=24m Theta=30 The range of projectile=u^2 x sin60/g 240=u^2 x √3/2 u^2 = 240 x 2/√3 u=4√30/√√3 Therefore required velocity=4√30/√√3 CHEEN TAPAK DUM DUM
initial velocity (u) is 20km/hr=5.55m/s Angle is 30° Gravity(g) is 9.8m/s^2 Range that is horizontal distance covered is, R=u^2xsin2(30°)/g =5.55x5.55x0.866/9.8 =2.72 Matlab thele par se gaadi sirf 2.72 meter aage jake giregi. Height attained by scooty is, H=u^2xsin^2(30°)/2g =5.55x5.55x0.25/19.6 =0.4 meter Matlab scooty Gaadi ko jake thokni chahiye. Time Required is, T=2usin(30°)/g =2x5.55x0.5/9.8 =0.6 seconds Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9
6:46 Velocity of body = 20m/s Angle of projection = 30degree gravity = 9.8m/s^2 Let R be the range R = v^2.sin2θ/g :. R = 35.3469 m So the scooter can travel a distance of 35.35 m The average length of the car is 4.5 m There are 6 cars. :. total distance = 27 m let each space = 1 m :. total space = 7 m (7 spaces. Let each car be in the middle of each space) :. Total distance needed to be covered = 34 m 35.35 m > 34 m left due to air resistance the distance is reduced by 1m :. distance covered = 34.35m So from the above numerical, we can prove that the scooter can cross the jam.
Given: - Mass of the object (m) = 350 kg - Initial speed (u) = 20 km/h - Angle of projection (θ) = 30° First, we need to convert the initial speed from km/h to m/s: 1 km/h = 1000 m / 3600 s = 5/18 m/s So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s Now, we can calculate the horizontal and vertical components of the object's velocity: Horizontal component (ux) = u * cos(θ) Vertical component (uy) = u * sin(θ) ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s Next, we can calculate the time of flight (t) using the vertical motion: The time to reach the highest point (time to peak) is given by: t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity) t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s Since the motion is symmetrical, the total time of flight is twice the time to reach the peak: t = 2 * 0.283 s = 0.566 s Finally, we can find the horizontal distance (range) the object will travel: Range = ux * t Range = 4.81 m/s * 0.566 s ≈ 2.72 meters Therefore, the object will land approximately 2.72 meters away from the point where it was launched.
6:33 Initial velocity 20km/h =20x5/18 = 5.5m/s Slope 30° from horizontal axis , varticall velocity hogi usinthetha= 2.75m/s and horizontal velocity hogi ucosthetha =4.8 m/s Ab hamare pass x and y component aa chuke h to easy rahega time of flight niklna. Let's find time of flight= 2u/g = 0.55 s Yaha pr maine Y component ki velocity li h maine cauz (TOF) vertical velocity pr depend krta h na ki horizontal pr. Range nikalte h aab, R=UxT= 4.8x0.55 =2.64m pr jakr land krega scooter Isme horizontal velocity li h cuz range horizontal velocity pr depend krta h.
Oo Bhai gazab sabke answer dekhe Maine sale sab log sidha formula dal ke answer bata rhe the lekin tumne to step wise pura solve kia h and explanation bhi accha kia h jaise vertical velocity hi kyu li time of flight mai 👍
6:15 In this Case
Let initial velocity (u) is 20km/hr=5.55m/s
Angle is 30°
Gravity(g) is 9.8m/s^2
Range that is horizontal distance covered is,
R=u^2xsin2(30°)/g
=5.55x5.55x0.866/9.8
=2.72
Matlab thele par se gaadi sirf 2.72 meter aage jake giregi.
Height attained by scooty is,
H=u^2xsin^2(30°)/2g
=5.55x5.55x0.25/19.6
=0.4 meter
Matlab scooty Gaadi ko jake thokni chahiye.
Time Required is,
T=2usin(30°)/g
=2x5.55x0.5/9.8
=0.6 seconds
Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9 seconds💀💀
(1 like to banata hai)
Edit: Nishchay Bhai please ♥️
Ooo BC mujhe tution dede vro
Jee aspirant spotted👍👍
@@Swara_47 I am preparing NEET😅😅
@@Harshvardhan_Patil108 Ohh! Sorry but atleast I was right that you are preparing for a competitive exam ! 😃
💀🫥🙏ram ram bhaiya 🫥🙏
6:41
Given:
- Mass of the object (m) = 350 kg
- Initial speed (u) = 20 km/h
- Angle of projection (θ) = 30°
First, we need to convert the initial speed from km/h to m/s:
1 km/h = 1000 m / 3600 s = 5/18 m/s
So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s
Now, we can calculate the horizontal and vertical components of the object's velocity:
Horizontal component (ux) = u * cos(θ)
Vertical component (uy) = u * sin(θ)
ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s
uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s
Next, we can calculate the time of flight (t) using the vertical motion:
The time to reach the highest point (time to peak) is given by:
t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity)
t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s
Since the motion is symmetrical, the total time of flight is twice the time to reach the peak:
t = 2 * 0.283 s = 0.566 s
Finally, we can find the horizontal distance (range) the object will travel:
Range = ux * t
Range = 4.81 m/s * 0.566 s ≈ 2.72 meters
Therefore, the object will land approximately 2.72 meters away from the point where it was launched.
Answer: The object will land approximately 2.72 meters away from the launch point.
💀💀💀
Broo forget everything about the outer world 💀💀💀
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U r great!! 😂
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6:37 Velocity = 20 km/hr = 20×1000/3600 = 50/9 m/s = 5.55 m/s
Angle of elevation with the ground= 30°
Range= (5.55)²sin(2×30°)/(9.8)
≈ 2.72201 metres
(Its the maximum range, g is taken constant, and the mass is neglected. If mass is also taken into account then the range would be less than this)
Sch btana engineering kr rhi ho na😅
Simaraa😅😅
Lemda = 9.8 always remember it by oure science teacher 😊 hehehh by the way wow but in 50/9m/s exact mean I didn't understand!
Hey bro , can u say how much speed and elevation with the ground we need to cover the distance showed in that serial, plss
Sidha answer batana tha
Pura chapter hi pada diya 😂😂
1:31 Most hilarious 😂😂😂😂😂😂😂😂😆😆
10:20 bhai tum to pehle se hi hero material ho😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊❤
6:37 at this point you asked us a ques
Ans. u=20km/h=5.5m/s
g=10m/s*2
Range of a projectile= u^2*sin2theta/g
=5.5^2 x sin60/10
=3.025 x √3/2
=2.6165
Pls like, it took some effort to calculate ❤❤❤❤
Length of 1 car= 4.8m
Length of 5 car=24m
Theta=30
The range of projectile=u^2 x sin60/g
240=u^2 x √3/2
u^2 = 240 x 2/√3
u=4√30/√√3
Therefore required velocity=4√30/√√3
CHEEN TAPAK DUM DUM
😂😂
🎉🎉 we haven't started trigonometry in class yet so I won't understand the solution BTW nischay should give heart to ur comment
Thankyou guys for the appreciation
Bro kuch samajh nhi aya 😂😂 I am. Commerce student
WHAT bro you r genius btw I m only in sixth but that trigonometry should be like basic
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Why 69😂
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😂😂😂😂
13:28 papita ka pate ko lagakar face transplant😂 That make me so laugh 😂😂😂
10:20 hero toh ho hi aap 😊😊
I
Yeh me kehne hi wali thi fir yeh comment dikh gayi😅
Yes but no my brother
4:01 shaddi bhi ho gya bacha bhi ho gya 😂😂😂😂
4:20 Statue 😂😂
Love ur videos😂❤❤ . This 14 min of video made my day❤
That's great
9:39 When Nischay said daru piyagi 💀💀
Iykyk 💀💀
Mai yahi comment dudne aya tha😂😂
oo daru badnaam kardi
Macher ka pate bhar Aaraha hai
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10:16 was chikna triggu's supremacy!!❤️❤️
2:11 this is so funny 😂😂😂😂
12:18 bhaiya is literally crying for his favorite food "bhel puri"😂😂
Hi
2:50 Nischay's narration is epic 😂🔥
initial velocity (u) is 20km/hr=5.55m/s
Angle is 30°
Gravity(g) is 9.8m/s^2
Range that is horizontal distance covered is,
R=u^2xsin2(30°)/g
=5.55x5.55x0.866/9.8
=2.72
Matlab thele par se gaadi sirf 2.72 meter aage jake giregi.
Height attained by scooty is,
H=u^2xsin^2(30°)/2g
=5.55x5.55x0.25/19.6
=0.4 meter
Matlab scooty Gaadi ko jake thokni chahiye.
Time Required is,
T=2usin(30°)/g
=2x5.55x0.5/9.8
=0.6 seconds
Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9
5:03 main aise hi statue hokar phone pakad ke video dekh rahi thi but fir aapki video dekh rahi hu toh hasi toh aa hi jaata hai 😂😂😂
12:50 the mood swings bruhhhh 😂😂😂
14:21
Didn't realise how these 14 min passed, literally amazing video enjoyed it so much.💖💗
Nischay bhiyaa always hero metrial ♥️🔥🔥🔥🔥🔥🔥😎
5:31 kya pose me statue huye he😂😂
3:05 😂😂😂what a shaadii Bhai 😂😂😂Indian tv serials 🎉boom
5:16 Statue scene ft. Nonu 😂🔥
English or Spanish 😂
It's giving English or Spanish vibe 👽💀🤷😂
थैंक्स for 6 subscribor complet ❤❤🎉🎉🎉
Think you 104 Subscribers ❤
BRO IN SHOCK 😶😶IN 1SEAND 4:09
Vanshaj 😂😂
5:26
the power of statue 😂
6:46
Velocity of body = 20m/s
Angle of projection = 30degree
gravity = 9.8m/s^2
Let R be the range
R = v^2.sin2θ/g
:. R = 35.3469 m
So the scooter can travel a distance of 35.35 m
The average length of the car is 4.5 m
There are 6 cars.
:. total distance = 27 m
let each space = 1 m
:. total space = 7 m (7 spaces. Let each car be in the middle of each space)
:. Total distance needed to be covered = 34 m
35.35 m > 34 m
left due to air resistance the distance is reduced by 1m
:. distance covered = 34.35m
So from the above numerical, we can prove that the scooter can cross the jam.
Insane bro ,itni mehnat kyu Bhai 😅
JEE aspirants be like:
bhai 20km/hr speed ka kaha tha , 20m/s speed bohot jyada hoti hai which is 72km/hr
bakloli kar di bro wrong speed lol
@@omnimamba8611 also friction was ignored here especially when it was on the slanted surface
0:01 reminds me of old Ashish Chanchlani videos 😂❤
Kamla OP😂😂
Sameeee😢 nostalgia ❤
Same❤
0:07 hehe
Exactly 💯😂
Mane aj pahli dfa video dakhi trigger Insan ki mza agya qsm sy what a editing 😂mane aj sy pahly Q na dakhi😂🤌🏻 subscribe to bnta ha ab😂
0:57 Shuru ho gaya roast 😂🔥 with physics
Physics m b ek kar rakhi saloo me
POSE THORA CASUAL HE🌚🤧👊🏻 5:40
Tum kon ho
69 likes ☠️💀
@@painfulsadbrokensong3452Chup gawar
@@painfulsadbrokensong3452 insan human😂
sahdeep2005😂50
4:02 Ohhhh vanshaj bhaiyaaa..acting😂😂😂😂...Aap sach me papa ban gye unke😂😅
Papa nhi maama
Maa Kasam jisne like nhibkiya
Kash aaj mere 8 Subscribers hogayen
Hello
@@fahimbhai0068hhh
Given:
- Mass of the object (m) = 350 kg
- Initial speed (u) = 20 km/h
- Angle of projection (θ) = 30°
First, we need to convert the initial speed from km/h to m/s:
1 km/h = 1000 m / 3600 s = 5/18 m/s
So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s
Now, we can calculate the horizontal and vertical components of the object's velocity:
Horizontal component (ux) = u * cos(θ)
Vertical component (uy) = u * sin(θ)
ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s
uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s
Next, we can calculate the time of flight (t) using the vertical motion:
The time to reach the highest point (time to peak) is given by:
t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity)
t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s
Since the motion is symmetrical, the total time of flight is twice the time to reach the peak:
t = 2 * 0.283 s = 0.566 s
Finally, we can find the horizontal distance (range) the object will travel:
Range = ux * t
Range = 4.81 m/s * 0.566 s ≈ 2.72 meters
Therefore, the object will land approximately 2.72 meters away from the point where it was launched.
13:35 Mera bhi mann ho raha hai 😑😢😂😂
1:45 🌚daru was personal 🐸
6:25 .. when an engineer become roster 😂😂😂😂
Bhai those marzil and vanjaj ke scenes gave me nostalgic memories back ❤.....
Becuz I'm re-watching this video after completing gaming insaan ❤
1:35 Sellmon Boi Driving 🤘😂😂
8:46 Kya hi energy dali he yrr video me❤❤.....Serial ke scene se jyada to aapko dekh k maja aa raha he😅😂...BTW missing your streams😢😢😢
6:55 Omg😂😂😂❤
Funniest part and Funniest video ever😂😂😂❤❤
Bhaiyaa aaapki sb videos ko dekhkr bina hsse rha hi nhi jata😅😅😂😊
Happy birthday to triggered insaan bigest fan from Kolkata pls reply me ❤❤
11:22 Coco melon 😂
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Triggu bro ke ye roast videos dekh ke bade maje aate h magar yahi kam agar dosto ke sath movie dekhte waqt kar do to 😅 Haye tauba .
12:52 old laugh is back 😂
7:02
😂😂 ro rahe he sab
4:05 wah bhaiya kya video hai maja aa gaya 😂😂aise hin saste raho aur hame hasate raho...
I would suggest you watch chotya bayochi mothi swapana . Absolute chaos😂
4:32 Statue 😡🤣
3:21 Nation wants to know 😂😂
😅 If you saw it😁 you saw it😅 (headset in girls ears )
3:51 this scene 😂
❤❤❤❤❤❤😂😂😂😂😊😊😊😅😅😅😅😅😅😅❤❤❤❤❤❤❤❤ hai na 😹 hai 😂
07:28 😂😂😂 ooh ohh ohho ohhoo 😂😂
6:33
Initial velocity 20km/h =20x5/18 = 5.5m/s
Slope 30° from horizontal axis , varticall velocity hogi usinthetha= 2.75m/s and horizontal velocity hogi ucosthetha =4.8 m/s
Ab hamare pass x and y component aa chuke h to easy rahega time of flight niklna.
Let's find time of flight= 2u/g = 0.55 s
Yaha pr maine Y component ki velocity li h maine cauz (TOF) vertical velocity pr depend krta h na ki horizontal pr.
Range nikalte h aab,
R=UxT= 4.8x0.55 =2.64m pr jakr land krega scooter
Isme horizontal velocity li h cuz range horizontal velocity pr depend krta h.
Oo Bhai gazab sabke answer dekhe Maine sale sab log sidha formula dal ke answer bata rhe the lekin tumne to step wise pura solve kia h and explanation bhi accha kia h jaise vertical velocity hi kyu li time of flight mai 👍
🏅🏆 Le bhai rakh le
you've absolutely smashed it mate.
🤯
Mind blowing sir g 🤓 konsi class mai ho aap 😭
9:19 logic 😂😂 rip
3:56 was epic finally nishchay ki shadi to hui chahe noni banake hui😂😂
Tum kon ho 😮
In dramo sa kafi achi apki video Hoti Hain bro,🥰🥰🥰
1:35 kya khatarnak roast krte ho nonu bhaiya😂😂
4:07 that 'mera lalla mera pati'🤣🤣🤣🤣
12:48 inki hasi🤣🤣🤣🤣
1:37 😂😂😂
10:15 style walk❌cute and crazy walk✔️
Bro u takes smile on my face at that time when I am. Weeping....
1:05 just nischay finding logics 😂😂
❤10:43 be like " Mujhe kiyu toda "😂😂😂😂
2:45 ap janata ko murk samjhna band kare ...hagu😂😂❤
Tum kon ho 😮
6:00 gravity ki bizti Kar di bhai😅😊😂
5:09 bro is still living in childhood😭😭✋🗿
12:57 Cry wala laugh ya laugh wala cry 😭🤣
12:19 ,literally 😂i am watching this video while eating bhel puri 😄
Me apki sari video dekhta hu .........koi ASI video nahi hai gispar me nahi hasa 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
Finally he made a Indian tv serials roast video after a long time😂
4:02 HE IS 😭😭✋
I CANT YRRR😂
He is who?
@@kamalnayan10thb42 hilarious..!!😭😭✋
@@Aryatiwari1709 He is who?
13:45 -- mujhe toh unse jada apne pe hasi a rahi hai
Pehle ye serial mai seriously dekhti thi😂
Pehle wale serial k naam kya tha?
Sath nibhana sathiya@@RG-lo8wr
@@RG-lo8wr I wanna know the same
aapne statue bolato me phone dekh tehi statue hogae aap sach me bahat camedy karteho❤❤😊😊😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂👍👍👍😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
I will study one minute for every like this comment gets 😅
Great
Tum kon ho 😮
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Like beggars noob
Bhai ab padhke dikha😂
10:40 3 guy be like: mujy kiu toda😂😂❤
Tum kon ho 😮
They are me , triggered insaan and the person 😂😂
@@ShamimaSultana-vg6ix indian ho
@@painfulsadbrokensong3452 nehi ..
tum kahan rahti Ho
3:02 Bhagte bhagte phere le liye 😂🤦🏻
10:15 wow omha❤❤❤❤
9:06 sasural simar ka❌
Makkhi✔️
Makhija returns
Sasural makkhi ka😂😂
9:33 vo aapaki kaju katli he😂😂
Tum kon ho 😮
14:05 bichara doctor 😂
Bhaiya Aap To Aise hi hero material hai aapko yah Bal badhane ki aur do button kholne ki jarurat Nahin Hai❤❤❤❤❤❤❤❤❤❤
3:45 that watch given by Abu is permanent in Nonu bhaiya's hand . ❤❤❤
Theta 30⁰
Velocity 20 km/hours = 5.5 m/s
Range = u² sin2∅/g
Range = 1.5125 √3
=2.619 meter
Kya bhai
Ae baigan 👀🤣🤣
Bhai isko math bolte h na😮
@@LUCKY_GAMERRR😂😂😂
Bhai Bhai
1:58 Seriel wale be like humare yaha aise hi hota h😂🤣
The fact that nobody is talking abt the different ppl and the room😂
* I came here after watching the epic and legendary show *
4:19 Statueee😂💀
2:30 doctor's be like hum job chorde ?? 🤣
3:10 itna inko diya mne😂😂😂
3:56 😂😂😂 41:21 statue 😂😂 12:25 hachiwali seen bohat cute thi yaaaar🤓💞💞appki hachi bohat cute he 12:45 😂😂❤
6:20..... when your parents want you to become a engineer......
But you want to be a UA-camr😂😂😂
He is engineer broo😂
@@Islam_is_the_best_religion-y yes
@@Islam_is_the_best_religion-y what do you mean
@@clashsmith2219 I meant ki you replied my comment after so long...so 🤣 thought ki you r aspirant..that's why less available on phone.
10:05 i could practically smell the edits
Yeah, I know how shit smells.
Just like trash
12:11 nischay yeah dekhne ke baad bhelpuri khana mat chod dena 😂😂
14:11 funniest part😂😂😂