Question

From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engineering building is 35.0 m tall. what is the initial velocity of the ball?

Answer 1

The diagram shown below illustrates the problem.
v =  vertical upward launch velocity.
g = 9.8 m/s² is acceleration due to gravity.

When s = - 35m, the elapsed time is 4 s.
Therefore
(- 35 m) = (v m/s)*(4 s) - (1/2)*(9.8 m/s²)*(4 s)²
-35 = 4v - 78.4
4v = 78.4 - 35 = 43.4
  v = 10.85 m/s

Answer: 10.85 m/s

Answer 2

Equations of the vertical launch:

Vf = Vo - gt

y = yo + Vo*t - gt^2 / 2

Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2

=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s

The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

Answer: 11.25 m/s