What was the total time of fall?

[keweezz]

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]

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]

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?

i got 45.78 for h, but it didn't work out ;(