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 4.0 sec, how fast must the centerfielder's throw leave his hand if he makes the catch then immediately throws the ball 350 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) 88 ft/sec
B) 64 ft/sec
C) 108 ft/sec
D) 152 ft/sec

Question

Quizplus · Accepted Answer

To solve for the initial velocity of the centerfielder's throw, we need to use the formula:

d = v_i*t + 0.5*a*t^2

where d is the distance traveled (350 ft), v_i is the initial velocity, t is the time the ball is in the air, and a is the acceleration due to gravity (which we can assume is -32 ft/s^2 since we are only looking at the vertical motion).

We can rearrange the formula to solve for v_i:

v_i = (d - 0.5*a*t^2)/t

We know that the time it takes the runner to go from 3rd base to home plate is 4.0 sec, so we can use that as our value for t. We also know that the distance from 3rd base to home plate is 90 ft. Using this information, we can calculate the time it takes for the ball to reach home plate:

90 ft = 0.5*a*t^2
t^2 = 5.625
t = 2.37 sec

Now we can plug in our values for d, t, and a to solve for v_i:

v_i = (350 - 0.5*(-32)*(2.37)^2)/2.37
v_i = 108 ft/sec

Therefore, the centerfielder's throw must leave his hand at a speed of 108 ft/sec.

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