Question

Solo 189

Answer 1

Note: Since I could not understand the height it was dropped from I will be using R. You can substitute whatever height in place of R.

Let the up be increasing y and down be decreasing y. We know the acceleration in the y direction which is,

$\ddot{y}=-10\frac{m}{s^{2}}=-g$

I'm going to call it -g for simplicity.

Now integrating to get velocity,

$\dot{y}=-gt$

Notice since the ball was dropped from rest, so the constant of integration is 0.

Integrating once more to get position,

$\[y=R-\frac{1}{2}gt^{2}\]$

<Since R is the initial height, it is the constant of integration here>

To answer the question for what time will it hit the ground, setting y=0 and solving for t we get:

$\[t=\sqrt{2gR}\]$

Now to get the velocity at which it hit the ground, we plug in this value of t into our velocity equation:

$\dot{y}(t=\sqrt{2gR}\)=-g\sqrt{2gR}$

Hope this helps!