Question

A ball is projected upward at time t = 0.0 s, from a point on a roof 70 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is What is the velocity of the ball when it is above the ground? −89 m/s − 72 m/s − 36 m/s − 107 m/s − 54 m/s

Answer 1

Answer:

The velocity of a ball will be "-70.13 m/s".

Explanation:

The given values are:

u = 70 m

t = 0.0 s

g = a = -9.8 m/s²

s = -1 m

v = ?

As we know,

The equation of motion will be:

⇒  $v^2-u^2=2as$

On substituting the estimated values, we get

⇒  $v^2-(70)^2=2\times (-9.8)\times (-1)$

⇒  $v^2-4900=19.6$

⇒  $v^2=19.6+4900$

⇒  $v^2=4919.6$

⇒  $v=\sqrt{4919.6}$

⇒  $v=70.13 \ m/s$

In downward direction, it will be:

⇒  $v=-70.13 \ m/s$