Question

A ball is thrown vertically downward from the top of a 35.9-m tall building. The ball then passes the top of a window that is 10.7 m above the ground 2.00 s after being thrown.

What is the speed of the ball as it passes the top of the window?

The equation for the speed of the ball, from time t=0 when the ball is released, is

V = Vo + g t
Vo it the initial velocity of the ball. The distance it has travelled at time t, measured vertically downward, is
Y = Vo t + (1/2) g t^2

You know that Y = 25.2 m at t = 2.
Use that information and the last equation to determine Vo. Then use that Vo in the first equation to get the value of V when t = 2.

Answer 1

25.2=Vo(2)+.5(-9.8)(2^2)
25.2=Vo(2)-19.6
25.2+19.6=2Vo
45.1=2Vo
Vo=22.55m/s

Answer 2

V=Vo + gt
V= 22.55 +9.8*2
V= 42.15 m/s

Answer 3

Displacement should be in negative... as the ball thrown vertically downward.....

Answer 4

So we solve.... s=ut - 1/2gt^2..... put t=2, s=-25.2m
After obtaining the value of u...using 1st eqn of motion... v=u-gt
Then v is ur required answer