Question

A ball is kicked on a planet the size of Earth but with twice the acceleration due to gravity. If the maximum height of the ball's trajectory is 39 meters, how long does it take the ball to fall back to the planet from this maximum height?

Answer 1

Answer:

The value $t = 1.995 \ s$

Explanation:

From the question we are told that

The acceleration due to gravity is a = 2g

The maximum height is h= 39 m

Generally from the kinematic equation

             $s =  ut + \frac{1}{2} at^2$

Here u  is the velocity of the ball at maximum height before it start falling and the value is  0 m/s

So

            $39 =  0* t + \frac{1}{2} (2g)t^2$

            $39 = (9.8)t^2$

            $t =  \sqrt{3.9795918}$

$t = 1.995 \ s$