In a baseball game a base-runner can attempt to advance when a fielder catches a hit ball before it reaches the ground, so long as the runner doesn't leave base before the fielder catches the ball. The fielder then attempts to throw the ball to the base ahead of the runner. If a base-runner can run from 3rd base to home plate in 4.6 sec, how fast must the centerfielder's throw leave his hand if he makes the catch then immediately throws the ball 370 ft from home plate? Assume the base-runner leaves his base at the same instant the centerfielder throws the ball, and that the ball reaches the catcher standing over home plate at the same height from which it was thrown and at the same time the runner would reach the plate. [Hint: This is basically a problem of determining how much initial vertical velocity the ball needs to stay in the air long enough to reach the plate.]
A) 80 ft/sec
B) 74 ft/sec
C) 109 ft/sec
D) 154 ft/sec

Question

Quizplus · Accepted Answer

We can use the formula d = vt + 0.5at^2, where d is the distance (370 ft), a is gravity (-32.2 ft/sec^2), t is the time the ball is in the air, and v is the initial velocity of the ball. We know that the time the ball is in the air is 4.6 sec, because that is how long it takes the runner to reach home plate. We also know that the ball starts at ground level and ends at ground level, so the initial and final vertical velocities are the same.

Plugging in the values we know, we get:

370 = v(4.6) + 0.5(-32.2)(4.6)^2

Solving for v, we get:

v = 109.1 ft/sec

Therefore, the centerfielder's throw must leave his hand at a speed of 109 ft/sec in order to reach home plate at the same time as the runner.

In a Baseball Game a Base-Runner Can Attempt to Advance