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

The maximum height reached by the ball is determined as 31.89 m.

How to calculate the maximum height reached by the ball?

The maximum height reached by the ball is calculated by applying the following formula as shown below;

v² = u² - 2gh

where;

• v is the final velocity of the ball
• u is the initial velocity of the ball
• g is acceleration due to gravity
• h is the maximum height reached by the ball

At maximum height, the final velocity of the ball is zero.

0 = u² - 2gh

2gh = u²

h = u² / 2g

The maximum height reached by the ball is calculated as;

h = (25 m/s)² / ( 2 x 9.8 m/s²)

h = 31.89 m

Answer 2

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

Step-by-step 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
`(m)/(s^(2))`

and u = 25 m/s

and v = 0

therefore s =
`(u^(2))/(2a)`

so s = 31.88 meters