A batter hits a baseball 2.25 ft above the ground towards 3rd base. If the ball

Question

A batter hits a baseball 2.25 ft above the ground towards 3rd base. If the ball takes 0.65 s to get to the 3rd baseman (who is 95 ft from homeplate) and it passes just above the jumping 3rd baseman's glove, what is the initial velocity of the ball? Assume the 3rd baseman (with his glove on) can reach to a height of 9 ft, and can jump 34 inches off the ground. How many m/s @ how many degrees, with respect to the ground? If the left field wall is 361 ft away, and 9 feet tall, will this result in a home run?

Explanation / Answer

We will need to do a few conversions for this problem.

First, 2.25 feet = .6858 m

95 feet = 28.95 m

9 feet = 2.743 m

34 inches = .8636 m

361 feet = 110.03 m

For the first part, we can apply d = vt for the x distance to the third baseman

28.95 = (v)(.65)

v = 44.54 m/s (x velocity)

In the y direction apply d = vot + .5at^2

d = (2.743 + .8636 - .6858) = 2.9208 m

2.9208 = .65v + (.5)(-9.8)(.65)^2

v = 7.68 m/s (initial velocity in the y direction)

The net velocity is found using the pythagoream theorem

v^2 = (44.54)^2 + (7.68)^2

v = 45.2 m/s

The angle is found using the tangent function...

Tan (angle) = (7.68)/44.54)

Angle = 9.78 degrees above the horizontal

In summary, for the first part, the velocity = 45.2 m/s at 9.78 degrees above the horizontal

For the second part, we need to find out how long it take to make it to the left field wall using the velocity in the x direction, since it does not change

d = vt

110.03 = (44.54)(t)

t = 2.47 sec.

Then, in the y direction, will it be high enough?

Apply d = vot + .5at^2

d = (7.68)(2.47) + (.5)(-9.8)(2.47)^2

d = -19.7 m, so no, this will not result in a home run

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