Question

Suppose that the estimate of the distance that the ball moves while a player is throwing it is 2.5 feet -- that is, the distance from where the ball is when the player starts their throw until it leaves their hand. Calculate the magnitude of the average acceleration in m/s2m/s2 that the ball has while it is being thrown, as it moves from rest to the point where it leaves the player's hand.

Answer 1

This question is incomplete. the complete question is;

a) Suppose that the estimate of the maximum height that a player can throw a baseball is 25 ft. For how long after the ball leaves the player's hand does it return to the players hand?

b) Suppose that the estimate of the distance that the ball moves while a player is throwing it is 2.5 feet -- that is, the distance from where the ball is when the player starts their throw until it leaves their hand. Calculate the magnitude of the average acceleration in m/s2m/s2 that the ball has while it is being thrown, as it moves from rest to the point where it leaves the player's hand.

Answer:

a) Total time of Flight is 2t is 2.49 seconds

b) The magnitude of the average acceleration is 98.03 m/s²

Explanation:

Given the data in the question;

a)

Maximum height $h_{max}$ = 25 ft = (25 /3.281) = 7.62 m

For how long after the ball leaves the player's hand does it return to the players hand;

$h_{max}$  = $\frac{1}{2}$gt²

we know that g = 9.81 m/s²

so we substitute

7.62 m =  $\frac{1}{2}$ × 9.81 m/s² × t²

7.62 m = 4.905m/s² × t²

t² = 7.62 m / 4.905

t² = 1.5535

t = √1.5535

t = 1.246 s

So total time of Flight is 2t = 2 × 1.246 s = 2.49 seconds

b)

given that

distance = 2.5 ft = ( 2.5 / 3.281) = 0.762 m

V = ?

from the First Equation of Motion

v = u + at

0 = u - 9.81 × 1.246 s

u = 12.2232 m/s

so from the Third Equation of Motion : v² = u² + 2as

(12.2232 m/s)² = 0² + 2 × a × 0.762 m

149.4066  = 1.524a

a = 149.4066 / 1.524

a = 98.03 m/s²

Therefore, the magnitude of the average acceleration is 98.03 m/s²