A pitcher throws a baseball horizontally from the mound to home plate. The ball falls 0.987 m (3.24 ft) by the

Question

A pitcher throws a baseball horizontally from the mound to home plate. The ball falls 0.987 m (3.24 ft) by the time it reaches home plate 18.3 m (60 ft) away. How fast was the pitcher's pitch?

Answer 1

The time it takes the ball to fly from the mound to home is
18.3=v0*t
t=18.3/v0
and the fall time is
0.987=.5*g*t^2
plug in t from above and solve for v0

2*0.987/9.81=18.3^2/v0^2
v0=18.3*sqrt(9.81/(2*0.987)
=40.8 m/s
or about 91 mph
j

Answer 2

First we would desire to make certain how long the ball replaced into interior the air. The stress of gravity reasons concern to enhance up on the same fee interior the vertical direction no remember how briskly this is moving interior the horizontal direction (for the purpose of this question anyhow). Delta x = Displacement or distance traveled (to that end its incredibly only distance) Delta t or only t = Time a = acceleration (to that end gravity) First we can regulate our common device container equation: delta x = a million/2 a t^2 to make the equestion: t = sqrt((2*delta x) / a ) = sqrt((2*0.885)/9.8) = .424985s So now all of us understand that the ball replaced into in flight for .424985 seconds from the pitcher to the to the dwelling house plate. The formula for velocity is: delta x / delta t or displacement over the years wich leaves us with 18.3m/.424985s, which = 40 3.0603m/s or whilst rounded good 40 3.1m/s And in case you want it converted to MPH, you're able to do this for the period of your guy or woman :-) desire that facilitates good success in physics.