Question

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m from the ground and making an angle of 17.3 o with the horizontal. How high will the ball be when it gets to first base?

Answer 1

When the ball goes to first base it will be 4.23 m high.

Step-by-step explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

`t=(40.5)/(28.64)=1.41s`

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.