Question

A ballplayer catches a ball 3.2 s after throwing it vertically upward. With what speed did he throw it? What height did it reach?

Answer 1

It's easier to solve this problem by considering either it's rise into the air or its fall back down. The ball will take as much time to reach highest point as it will to fall from that point back to the ballplayer. So you can say it will take half the time given to reach the top and the other half to fall back. This time is half of 3.2 seconds which is 1.6 seconds. The speed of the ball when it leaves the player's hands will be the same as when it returns. We can find the speed and distance for either the up or down trip.

At the ball's highest point its speed is zero. Let's use this value as the initial speed(vi). The speed of the ball when it returns to the players hands is the final speed(vf) and is what we want to find. We also know that the acceleration of gravity is 9.8m/s^2 and we know the initial speed(vi) is zero. We can find vf using the equation

a)
vf = vi + a*t

or using g for gravitational acceleration

vf = vi + g*t

Substituting our known values for vi and g we get

vf = vi + g*t
vf = 0 + (9.8)*(1.6)
vf = 15.68 m/s <===Answer
He throws with 15.68 m/s

b)
We can find the distance traveled using

d = vi*t + 1/2g*t^2

Substituting our values for vi, g and t we get

d = 0*(1.6) + (1/2)*(9.8)*(1.6)^2

d = 12.54 m
Height is 12.54 m