A ball, dropped from rest, covers three-sevenths of the distance to the ground in the last two seconds of its fall.
(a) From what height was the ball dropped?
_m
(b) What was the total time of fall?
_s
For a, don't i need the final velocity to solve the problem? Vi=o A=-9.81?
2 days ago - 1 day left to answer.
HallsofIvy
Aug25-08, 04:03 PM
A ball, dropped from rest, covers three-sevenths of the distance to the ground in the last two seconds of its fall.
(a) From what height was the ball dropped?
_m
(b) What was the total time of fall?
_s
For a, don't i need the final velocity to solve the problem? Vi=o A=-9.81?
2 days ago - 1 day left to answer.
No, you don't need the final velocity (although you can calculate it after you have solved b). I presume you know that the height of the ball after t seconds is s= h- (1/2)At2= h- 4.905t2. It will hit the ground when s= h- 4.905t2= 0 so the time it hits the ground, T, is T= \sqrt{h/4.905}.
the ball "covers three-sevenths of the distance to the ground in the last two seconds of its fall." which tells us that s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h.
Solve that equation for h.
keweezz
Aug25-08, 05:12 PM
No, you don't need the final velocity (although you can calculate it after you have solved b). I presume you know that the height of the ball after t seconds is s= h- (1/2)At2= h- 4.905t2. It will hit the ground when s= h- 4.905t2= 0 so the time it hits the ground, T, is T= \sqrt{h/4.905}.
the ball "covers three-sevenths of the distance to the ground in the last two seconds of its fall." which tells us that s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h.
Solve that equation for h.
s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h. For that part, i just do h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h that and solve for h?