Question

A ball with an initial velocity of 25 m/s is subject to an acceleration of -9.8m/s^2 how high does it go before coming to a momentary stop ?

Answer 1

Answer:

The ball will reach an height of 31.88 meters before coming to a stop.

Explanation:

we know that by equation of motion $v^{2} = u^{2} + 2as$

where v is the final velocity of the ball , u is the initial velocity of the ball , a is acceleration due to gravity and s is the distance traveled by the ball.

a = -9.8 $\frac{m}{s^{2}}$

and u = 25 m/s

and v = 0

therefore s = $\frac{u^{2}}{2a}$

so s = 31.88 meters