Hi everyone, question can also be answered using free fall equation with zero initial velocity. The height is 308.2 m = (1/2) * g * t² and find t = 7.9 s to get down to the ground. Then add the other time t = 8.52 s to reach maximum height which gives the total time of flight 16.42 s.
Acceleration is not constant in this problem. There's a period when the rocket accelerates from its thrusters, and a period when the rocket coasts in free fall. You have to put these two periods together as a piecewise problem of two constant acceleration periods, to solve for the maximum height.
acceleration is 2m/s2 and gravity acceleration is 9.8 which is pulling the rocket downwards , why we are not doing 9.8-2 for resultant acceleration or 2 is the final acceleration considering 9.8 ?
It is given in this problem that the acceleration is 2 m/s^2 upward. How this is caused, is irrelevant to the problem. The thrusters will provide the upward force needed to both oppose gravity AND cause an acceleration of 2 m/s^2. So on net, the force per unit payload mass of the rocket, is 11.8 N/kg.
How do you aply this to gravity with respect to starting distance to find the velocity at the ending destance. This equasion should not include time, just distance from object. I can not use the equasion “square root of 2gh” because that is implying that the stringth of gravity is constant and does not change with respect to distance so that equasion does not work for me.
Near the Earth's surface, gravity might as well be uniform, and the equation ultimately turns into v=sqrt(2*g*h) when falling a distance of h. When h is in meters, and the radius of Earth is in megameters, the scale of h relative to the radius of Earth is insignificant, and the gravitational field is close enough to uniform that it makes no difference to account for gravity's variation with position. If you wanted to account for how gravity varies with position, then you'd use the equation GPE = -G*M*m/r, and construct two versions of this with r=R and r=R+h in each of them. So you'd get deltaGPE = -G*M*m*(1/(R+h) - 1/R, which reduces to deltaGPE=G*M*m*(1/R - 1/(R+h)). Equate to KE=1/2*m*v^2, the little m cancels, and we get v=sqrt(G*M*(1/R - 1/(R+h))). G= universal constant of gravitation M = mass of Earth R = initial radial position from center of Earth, not necessarily the radius of Earth.
These lectures are really ambitious.The explanations are so perfect and clear.thank you professor
Hi everyone, question can also be answered using free fall equation with zero initial velocity. The height is 308.2 m = (1/2) * g * t² and find t = 7.9 s to get down to the ground. Then add the other time t = 8.52 s to reach maximum height which gives the total time of flight 16.42 s.
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For c) can't we just use Y=Y0+V0t+1/2at^2 ? We know the variables and get Time from here and not calculate Velocity, that we are not asked for?
Acceleration is not constant in this problem. There's a period when the rocket accelerates from its thrusters, and a period when the rocket coasts in free fall. You have to put these two periods together as a piecewise problem of two constant acceleration periods, to solve for the maximum height.
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acceleration is 2m/s2 and gravity acceleration is 9.8 which is pulling the rocket downwards , why we are not doing 9.8-2 for resultant acceleration or 2 is the final acceleration considering 9.8 ?
Gravity acceleration is (-9.81)
It is given in this problem that the acceleration is 2 m/s^2 upward. How this is caused, is irrelevant to the problem. The thrusters will provide the upward force needed to both oppose gravity AND cause an acceleration of 2 m/s^2. So on net, the force per unit payload mass of the rocket, is 11.8 N/kg.
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Nu complicated rocket designs to ascend a new electron into the ionosphere
How do you aply this to gravity with respect to starting distance to find the velocity at the ending destance. This equasion should not include time, just distance from object. I can not use the equasion “square root of 2gh” because that is implying that the stringth of gravity is constant and does not change with respect to distance so that equasion does not work for me.
Near the Earth's surface, gravity might as well be uniform, and the equation ultimately turns into v=sqrt(2*g*h) when falling a distance of h. When h is in meters, and the radius of Earth is in megameters, the scale of h relative to the radius of Earth is insignificant, and the gravitational field is close enough to uniform that it makes no difference to account for gravity's variation with position.
If you wanted to account for how gravity varies with position, then you'd use the equation GPE = -G*M*m/r, and construct two versions of this with r=R and r=R+h in each of them. So you'd get deltaGPE = -G*M*m*(1/(R+h) - 1/R, which reduces to deltaGPE=G*M*m*(1/R - 1/(R+h)). Equate to KE=1/2*m*v^2, the little m cancels, and we get v=sqrt(G*M*(1/R - 1/(R+h))).
G= universal constant of gravitation
M = mass of Earth
R = initial radial position from center of Earth, not necessarily the radius of Earth.
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I run my own site www.MathAndScience.com. Thanks! Jason
Why was the height negative
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