Question

How do this question? about Vertical launch

From the top of a building of 30m, was thrown up an object that hit the ground 4s after it was released. Neglecting the resistance of the air and adopting g = 10m / s², calculate:
A) The speed at which the object was launched
B) the maximum height reached by the object in relation to the ground
C) the speed of the object upon reaching the ground
D) the hourly functions S (t) and V (t) of the movement developed by the object

Answer 1

It's a vertical launch so the `x` component of velocity and position remain zero.

`v_y = v _ 0 - g t`

`y = 30 + v_0 t - 1/ 2 g t^2`

We set the initial `y` to `30` indicating a height of `30` when `t=0`.

We're told when `t=4` we get `y=0`.

`0 = 30 + v_0 (4) - 1/2 g (4)^2`

`v_0 = 1/4(g/2 (4)^2- 30) = 50/4 = 12.5` m/s `quad` Answer for (A)

We have max height when `v_y=0,` i.e.

`t= 12.5/g = 1.25 sec`

`y_text{max} = 30 + 12.5 (1.25) - 1/2 (10)(1.25)^2 = 37.8125` m `quad` For (B)

Speed at `t=4`

`v(4) = v _ 0 - g t = 12.5 - 10(4) = - 27.5` m/s `quad` For (C)

For (D) I have no idea what an hourly function is. We found

`y(t) = 30 + 12.5 t - 5 t^2`

`v(t) = 12.5 - 10 t`