Question

A softball is hit over a third baseman's head with some speed v0 at an angle θ above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V=7.00m/s. He catches the ball t=2.00s later at the same height at which it left the bat. The third baseman was originally standing L=18.0m from the location at which the ball was hit.
A) Find v0. Use g=9.81m/s2 for the magnitude of the acceleration due to gravity.

Express the initial speed numerically in units of meters per second to three significant figures.

B) Find the angle θ in degrees.

Express your answer numerically in degrees to three significant figures.

Answer 1

Answer:

A) v0 = 18.8 m/s

B) θ = 31.5°

Explanation:

On the x-axis:

$X_{ball} = X_{player}$

$v0*cos\theta*t=L+V*t$

where

t = 2s;  L = 18m;  V = 7m/s

$v0*cos\theta*2=32$

$v0=16/cos\theta$

On the y-axis for the ball:

$\Delta Y=v0*sin\theta*t-1/2*g*t^2$

$0 = v0*sin\theta*2-19.62$

Replacing v0:

$0 = tan\theta*2*16-19.62$

$\theta=atan(19.62/32)=31.5\°$

Now, the speed v0 was:

v0 = 18.8m/s