A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

hamzaahmeds hamzaahmeds · Answer 2 · 2021-09-03T20:03:50+02:00

This question deals with the concept of semi-projectile motion. It can be solved using the equations of motion in both horizontal and vertical directions.

The ball falls a distance of 1.03 ft, vertically.

First, we will consider the horizontal motion of the ball. The air resistance is assumed negligible. Therefore, the horizontal velocity of the ball remains constant. So, we can use the formula for constant velocity here:

s = vt

where,

s = horizontal distance covered to reach home plate = 60.5 ft

v = speed of the ball = (90 mi/h)(5280 ft/ 1 mi)(1 h/3600 s) = 132 ft/s

Therefore,

60.5 ft = (132 ft/s)t

t = 0.46 s

Now, we apply the second equation of motion to the vertical motion:

[tex]h = v_it +\frac{1}{2}gt^2[/tex]

where,

h = vertical distance the ball fell = ?

[tex]v_i[/tex] = initial vertical velocity = 0 m/s (since the ball was thrown horizontally)

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]h = (0\ m/s)(0.46\ s)+\frac{1}{2} (9.81\ m/s^2)(0.46\ s)^2[/tex]

h = 1.03 ft

The attached picture shows the equations used in semi-projectile motion.

Learn more about projectile motion here:

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