A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground. a. How high does the ball rise?
b. How long does the ball remain in the air? Hint: The time it takes the ball to rise equals the time it takes to fall.

a. 25.80 m

b. 4.59s

Explanation:

(refer to attached for reference)

Part a :

We are asked to find the distance the ball rises "s"

The relevant equation of motion is:

v² = u² + 2as,

Given:

Initial speed, u = 22.5 m/s

As the ball rises, we can also deduce the following:

1) the ball rises to a maximum height before it starts falling again. At the point of maximum height, for a split second, speed of the ball is zero.

Hence we can deduce that final speed, v = 0 m/s

2) the only acceleration acting on the ball as it travels is gravity acting downward. Hence the acceleration that the ball experiences is equal to negative gravitational acceleration, i.e a = - 9.81 m/s²

Simply substitute the above values into the equation above and solve for distance "s"

v² = u² + 2as

0² = 22.5² + 2(-9.81)s

19.62s = 22.5²

s = 22.5² / 19.62 = 25.80 m (answer)

Part b:

We are asked to find how long the ball takes to rise and fall again before it is caught. We are also told that the time for the ball to rise is the same as the time for the ball to fall. Hence all we need to do is find the time taken for the ball to rise and then multiply that by 2 to get our answer.

In this case, the relevant equation of motion is:

v = u + at, where t is the time taken for the ball to rise.

simple substitute the above known values into this equation and solve for t

v = u + at

0 = 22.5 + (-9.81)t

9.81t = 22.5

t = 22.5/9.81

t = 2.29 s

hence the time taken to rise AND fall = 2t = 2(2.29) = 4.59s (answer)

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