A 8kg ball is thrown up with an initial speed of 20m/s. Suppose there is a constant air resistence of 30N opposing the motion during its entire up and down path. How long will it take to get to the highest point in its path? how high is the highest point? How much work was done by the air resistence during the entire path up and down? What is the kinectic energy when it gets back to its original point?
6 answers
I will be glad to critique your work.
I don't how to start it off
When the ball reaches its highest point H, potential energy will be
M g H = (1/2)MVo^2 - f*H
where f is the friction force, 30N.
H = (1/2)MVo^2/(Mg + f)
Solve for H.
The time to get to that high point will be
T = H/Vo, since it will be decelerating at a constant rate.
The KE when it hits the ground will be M g H - f*H. This will equal
(1/2)MVo^2 - 2fH
M g H = (1/2)MVo^2 - f*H
where f is the friction force, 30N.
H = (1/2)MVo^2/(Mg + f)
Solve for H.
The time to get to that high point will be
T = H/Vo, since it will be decelerating at a constant rate.
The KE when it hits the ground will be M g H - f*H. This will equal
(1/2)MVo^2 - 2fH
Thank you. one more question what would be the acceleration on the way up? on the way down?
acceleration due to gravity
The acceleration is
g + (f/M) on the way up, and g - (f/M) on the way down. It is pointed down in both cases.
They are different because both g and friction point down on the way up, but are in opposite directions on the way down. For that reason, it arrives at the ground at a lower velocity than it took off.
g + (f/M) on the way up, and g - (f/M) on the way down. It is pointed down in both cases.
They are different because both g and friction point down on the way up, but are in opposite directions on the way down. For that reason, it arrives at the ground at a lower velocity than it took off.
1 answer