Question

A major league pitcher stands on the pitching mound 60ft 6in from home plate throws a fastball at 105 mph. Assume both the batter and pitcher are the same height (6ft 6in) and the ball is released 55 inches above the ground when thrown horizontal.

1. How far will a 340 gram baseball travel before hitting the ground?
2. How high is the ball above the ground when it passes over home plate?
3. How long does it take to reach home plate?

Now let’s look at what happens when the ball is hit, using the setup from the last problem. When hit the ball’s speed is an average of 30% faster and leaves the bat at a 5 degree angle.

4. How high is the ball when it passes the pitching mound?
5. How far will the ball travel?

Answer 1

Solution

a. Velocity of the baseball = 105 miles per hour = 105 * 1.6 * 1000/3600 m/s = 46.67 m/s

Height of the ball when it is thrown horizontally = 55 in + 6 ft 6 in = 1.4 m + 1.8 m + 0.15 m = 3.35 m

Because the pitcher has not applied any vertical forces to the ball, the only force affecting the ball is gravity. This means the ball takes just as long to fall to the ground as it would if it were dropped, despite it now travelling a large horizontal distance in the duration.

Let us find out the the time taken by the ball to hit the ground using the equation

s = ut + (1/2)at²

s= 3.35 m

u = 0 m/ s (because initially no vertical force is applied to the ball)

a = g = 9.8 m/s²

3.35 = 0 + (1/2) * 9.8 * t²

t² = 0.68

velocity of the ball in the horizontal direction = 46.67 m/s

distance travelled by the ball before hitting the ground, d = vt = 46.67 * 0.83 = 38.74 m

b. Distance of the home plate from the pitching mound = 60 ft 6 in = 18.44 m

Time taken by the ball to reach this distance, t = d/v = 18.44/46.67 =  0.4 s

Again using the equation, s = ut + (1/2) at², this time to find the distance s, we get

s = 0 + (1/2) * 9.8 * (0.4)² = 0.78 m

Height of the ball above the ground when it reaches the home plate = 3.35 - 0.78 = 2.57 m

c. We already found out above the time taken by the ball to reach the home plate, t = 0.4 s

d. Horizontal velocity of the ball when it reaches the home plate will be the same, i.e, 46.67m/s

The vertical velocity of the ball when it reaches the home plate can be found out using the equation,

v² = u² + 2as

u = 0

a = g = 9.8 m/s

s = 18.44 m

v² = 2 * 9.8 * 18.44 = 361.424

the total final velocity,

130 % of v_f = (130/100) * 50.39 = 65.5 m/s

Angle at which ball is hit = 5^o

Vertical component of velocity = 65.5 * sin 5⁰ = 5.70 m/s

time taken to reach the horizontal distance 46.67 m = 46.67/ (65.5 * cos 5⁰) = 0.72 s

Using the formula s = ut + (1/2)at²,

s = 5.7 * 0.72 + (1/2) * -9.8 * 0.72² = 1.56 m

e. Vertical component of the velocity when the ball falls in the ground = -5.70 m/s

final velocity - initial velocity = acceleration * time

-11.4 = -9.8 * time

time = 1.16 s

Distance travelled by the ball before falling into the ground = horizontal velocity * time = 65.5 * (sin 5^o) * 1.16 = 75.22 m