5:56 would Vavg really change if air resistance was taken into account? Because Vi = 49m/s and Vf = 0m/s still? I think the time will be shorter which affects max height, but the Vavg would stay the same. Although, the Force of air resistance would be greater at the higher speeds, so maybe the ball would never reach its initial velocity on the way down? If that is true, then the Vavg would only be lower on the way down?
You're right, average velocity wouldn't change on the way up, but the time of flight would be shorter so the max height would be lower. And you're right yet again that on the way down the final speed downward will be lower than the initial speed upward which would lower the average velocity on the way down. Nice!
At 17:28 how come adding 49 + 84.9 will give 98 m/s. I'm confused about finding the final velocity. I know the final is equal to the initial but adding those doesn't give me 98... can you explain, please?
Around 9:02 when you say that there is only acceleration in the vertical component and none in the horizontal direction, does that apply to all projectile motion questions or specifically just the one mentioned in the video?
It applies to most of the examples you are likely to encounter like throwing a stone or kicking a ball. The key is that there are no forces operating in the x-direction; gravity only operates in the y-direction (as long as we're on the surface of the earth anyway). But if we actually take into account air resistance (which we typically don't) it points in the direction opposite of the motion and will therefore typically have an x-compoent. Or if we had a little toy rocket with an engine it could still have a force operating in the x-direction. But these are not the types of examples you're likely to encounter when studying projectile motion. Hope this helps!
Sorry Casey as I totally skipped a step and made an assumption here. But the initial velocity in the y-direction was 49m/s and the final velocity in the y-direction at its max height was zero and I simply took the average of 49m/s and zero (49 + 0)/2. Hope this helps!
At 5:00 you said we could use the first equation, with using it, I got 375m. Can you explain how we could've used this equation to get the correct answer?
oh my gosh, i think i'll actually pass the 8 week physics class I accidentally took instead of 16 weeks. Thank you!!
You're welcome - I hope you do well!
5:56 would Vavg really change if air resistance was taken into account? Because Vi = 49m/s and Vf = 0m/s still? I think the time will be shorter which affects max height, but the Vavg would stay the same.
Although, the Force of air resistance would be greater at the higher speeds, so maybe the ball would never reach its initial velocity on the way down? If that is true, then the Vavg would only be lower on the way down?
You're right, average velocity wouldn't change on the way up, but the time of flight would be shorter so the max height would be lower. And you're right yet again that on the way down the final speed downward will be lower than the initial speed upward which would lower the average velocity on the way down. Nice!
At 17:28 how come adding 49 + 84.9 will give 98 m/s. I'm confused about finding the final velocity. I know the final is equal to the initial but adding those doesn't give me 98... can you explain, please?
Pythagoras theorem - sqrt (49^2 + 84.9^2) gives you the 98m/s
Around 9:02 when you say that there is only acceleration in the vertical component and none in the horizontal direction, does that apply to all projectile motion questions or specifically just the one mentioned in the video?
It applies to most of the examples you are likely to encounter like throwing a stone or kicking a ball. The key is that there are no forces operating in the x-direction; gravity only operates in the y-direction (as long as we're on the surface of the earth anyway). But if we actually take into account air resistance (which we typically don't) it points in the direction opposite of the motion and will therefore typically have an x-compoent. Or if we had a little toy rocket with an engine it could still have a force operating in the x-direction. But these are not the types of examples you're likely to encounter when studying projectile motion.
Hope this helps!
@@ChadsPrep That definitely helps a lot Chad! Thank you so much for these videos :)
@@0RANGOTANG You're welcome!
15:26 how did you get average velocity?
Sorry Casey as I totally skipped a step and made an assumption here. But the initial velocity in the y-direction was 49m/s and the final velocity in the y-direction at its max height was zero and I simply took the average of 49m/s and zero (49 + 0)/2. Hope this helps!
At 5:00 you said we could use the first equation, with using it, I got 375m. Can you explain how we could've used this equation to get the correct answer?
I will say using the equation: Vf^2=Vo^2+2a (delta x) I got 125m
I see where I messed up. 1. I didnt multiple t^2 by a. 2. I made a +10 instead of -10 . Fixing my mistakes, I got 125m. thank you
Awesome work Lexxus - keep it up :)