Question

if the height from which a body is released is numerically equal to the velocity acquired finally then the value of that height is ​

Answer 1

Explanation:

highet is u^2sin^2 theta/2g

Answer 2

Given: u=0, g=10m/s^2

Let height be h and final velocity be v.

h=v (Given)

Using 3rd equation of motion:

${v}^{2} = {u}^{2} + 2gh$

${v}^{2} = 0 + 2 \times 10 \times h$

${v}^{2} = 20v$

${v}^{2} - 20v = 0$

$v(v - 20) = 0$

$v = 0 \\ or \\ v = 20$

Since height/velocity cannot be 0

$v = h = 20$