Question

Assuming constant velocities, if a fastball pitch is thrown and travels at 40 m/s toward home plate, 18 m away, and the head of the bat is simultaneously traveling toward the ball at 18.0 m/s, how much time elapses before the bat hits the ball?1) About 0.3 s2) About 0.6 s3) About 0.9 s4) About 1.2 s

Answer 1

Answer:

1) About 0.3 s

Explanation:

The ball is thrown at 40 m/s and the bat is simultaneously traveling toward the ball at 18 m/s, then, the ball and the bat are getting closer at 58 m/s because:

40 m/s + 18 m/s = 58 m/s

Additionally, they need to travel 18 m, so replacing d by 18 m and v by 58 m/s we can calculate the time to hit the ball as:

$t=\frac{d}{v}=\frac{18m}{58\text{ m/s}}=0.3\text{ s}$

Therefore, the answer is:

1) About 0.3 s