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.

a. How far will a 340 gram baseball travel before hitting the ground?

b. How high is the ball above the ground when it passes over home plate?

c. 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.

(a) How high is the ball when it passes the pitching mound?

(b) How far will the ball travel?

Answer 1

1.a)

Given,

The speed of a baseball, v = 105 mph

= 105 * 1.6 * 1000 / 3600 = 46.67 m/s

The height if the ball when horizontally thrown, h = 55 in + 6 ft 6 in h = 1.397 m + 1.98 m = 3.38 m

By using the equation,

s = ut + (1/2) at²

s = 3.38 m, u = 0 m/s, a = 9.8 m/s²,

3.38 = 0 + 0.5 * 9.8 * t²

t² = 0.689

t = 0.83 sec

The distance travelled by the ball before hitting the ground, d = vt

d = 46.67 * 0.83 = 38.74 m

b)

The distance of the home plate from the pitch in mound, d = 60 ft 6 in

d = 18.47 m

The time taken by the ball to reach this distance is, t = d / v

t = 18.47 / 46.67

= 0.396 sec

By using,

s = ut + (1/2) a t²

s = 0 + 0.5 * 9.8 * 0.396²

= 0.768 m

The height of the ball above the ground when it reaches the home plate = 3.38 - 0.768

= 2.612 m

c)

The time taken by the ball to reach the home plate is, t = 18.47 / 46.67

t = 0.396 sec

2.a)

The horizontal velocity of the ball when it reaches the home plate is the same. So, v = 46.67 m/s

The vertical velocity of the ball when it reaches the home plate is obtained by using,

v² - u² = 2as

v² = 2as ( u = 0 m/s)

v² = 2 * 9.8 * 18.47

v = 19.03 m/s

The final velocity, v_f = √ ( 46.67)² + (19.03)²

v_f = 50.4 m/s

30 % of v_f = (30 / 100) * 50.4

= 15.12 m/s

Angle, = 5^o

The vertical component of velocity, u = 15.12 * sin 5⁰ = 1.32 m/s

The time taken to reach the horizontal distance, t = 46.67/ (15.12 * cos 5⁰) = 3.1 s

By using the formula,

s = ut + (1/2)at²,

s = 1.32 * 3.1+ 0.5 * (- 9.8) * 3.1²

= 4.1 - 47.1

= 43 m