Question

A pitcher throws a fastball at 140 km/hr toward home plate, which is 18.4 m away. Neglecting air resistance (not a good idea if you are the batter), find how far the ball drops because of gravity by the time it reaches home plate.

Answer 1

Step-by-step explanation:

Assuming the ball is thrown horizontally and has no 'English' :

140 km/hr = 38.89 m/s

Time to reach home plate : distance / rate = time

18.4 m / 38.89 m/s = .4731 s

Vertical distance = 1/2 a t^2

= (1/2) (9.81)(.4731)^2 =~ 1.1 m drop

Answer 2

The ball drops approximately 0.106 meters (or 10.6 centimeters) due to gravity by the time it reaches home plate.

Step-by-step explanation:

The horizontal distance that the ball drops due to gravity can be calculated using the equation d = ½gt², where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight. To find the time of flight, we can use the formula t = d/v, where v is the horizontal velocity of the ball. In this case, the horizontal velocity is 140 km/hr, which is equivalent to 38.9 m/s. Plugging the values into the equation, we have:

t = 18.4 m / 38.9 m/s = 0.473 s

Now we can calculate the vertical distance that the ball drops using the formula h = (1/2)gt².

h = (1/2)(9.8 m/s²)(0.473 s)² = 0.106 m

Therefore, the ball drops approximately 0.106 meters (or 10.6 centimeters) due to gravity by the time it reaches home plate.