A baseball pitcher throws a ball with a maximum speed of 150. km/h.

If the ball is thrown horizontally, how far does it fall vertically by the time it reaches the catcher's glove 20. m away? Give your answer numerically in meters using decimal notation (omit the units).

Use the acceleration due to gravity as g=9.8 m/s2

Ignoring air resistance (which is considerable in this case of 150 km/h)

Convert the speed into m and s.
150 km/s
= 150000/3600 m/s
=41.667 m/s
Horizontal distance
=20 m
Time required, t
=20/41.678 s
=0.48 s.
The ball was thrown horizontally, so the initial vertical velocity, u=0 m/s
Acceleration, a = -g = -9.8 m/s/s
Vertical distance
=ut + (1/2)at²
=0*0.48 + (1/2)*(-9.8)0.48²
=-1.129 m.

I assume you have a computer to check the answer, so determine the number of decimal places you need to present, 1.1, or 1.13, or 1.129, or in negative (downwards).

To solve this problem, we need to calculate the vertical distance the ball falls by the time it reaches the catcher's glove.

First, we need to determine the time it takes for the ball to travel the horizontal distance of 20 m. Since the ball is thrown horizontally, we can ignore the initial vertical velocity and consider only the horizontal velocity. The formula for calculating time is:

time = distance / velocity

Plugging in the given values, we get:

time = 20 m / 150 km/h

To convert kilometers per hour (km/h) to meters per second (m/s), we need to multiply by 1000/3600:

time = 20 m / (150 km/h) * (1000 m/1 km) * (1 h/3600 s)
= 20 / (150 * 1000 / 3600)

Simplifying the expression:

time = 20 / (15 * 1000 / 36)
= 20 * (36 / (15 * 1000))
= 20 * (36 / 15000)
= 20 * 0.0024

Calculating further:

time = 0.048 seconds (rounded to three decimal places)

Now that we know the time it takes for the ball to reach the catcher's glove, we can calculate the vertical distance the ball falls using the formula:

distance = (1/2) * g * time^2

Plugging in the values:

distance = (1/2) * 9.8 m/s^2 * (0.048 s)^2
≈ 0.011 m

Therefore, the ball falls vertically by approximately 0.011 meters when it reaches the catcher's glove 20 meters away.