Question

a body dropped from a height reaches a velocity of 13m/s just before touching the ground. What is the initial height of the ball and the time of height​

Answer 1

Hi there!

We can use the following (derived) equation to solve for the final velocity given height:

vf = √2gh

We can rearrange to solve for height:

vf² = 2gh

vf²/2g = h

Plug in the given values (g = 9.81 m/s²)

(13)²/2(9.81) = 8.614 m

We can calculate time using the equation:

vf = vi + at, where:

vi = initial velocity (since dropped from rest, = 0 m/s)

a = acceleration (in this instance, due to gravity)

Plug in values:

13 = at

13/a = t

13/9.81 = 1.325 sec