Explanation:
It is given that,
A ball hits the ground at 40 m/s
Let h is the height from where you drop a ball. It is based on the conservation of energy as :
[tex]mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\\text{Putting all the values in above formula}\\\\h=\dfrac{40^2}{2\times 9.8}\\\\h=81.63\ m[/tex]
It is dropped from a height of 81.63 m.
Let it take t seconds to hit the ground. When it hits the ground, its final speed, v = 0 and u = 40 m/s. So,
[tex]h=ut+\dfrac{1}{2}gt^2\\\\h=\dfrac{1}{2}gt^2\\\\t=(\dfrac{2h}{g})^{1/2}\\\\t=(\dfrac{2\times 81.63}{9.8})^{1/2}\\\\t=4.08\ s[/tex]
So, it will take 4.08 seconds to hit the ground.
