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?

Question

Physics

Ricky Bates

26 July, 11:04

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?

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Answers (2)

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Maria Burnett · Answer 1

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

Casey · Answer 2

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