A major-league pitcher can throw a baseball in excess of 46.2 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 18.9 m away from the point of release?
time to travel: t=18.9/46.2
distance fallen: 4.9 t^2
To find out how much the ball will drop by the time it reaches the catcher, we can make use of the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
where:
y = vertical displacement of the ball
y0 = initial vertical position of the ball (which is 0 in this case since it's thrown horizontally)
v0y = initial vertical velocity of the ball (which is 0 since it's thrown horizontally)
g = acceleration due to gravity
t = time
Since the ball is thrown horizontally, its initial vertical velocity (v0y) is 0. Therefore, the equation reduces to:
y = - (1/2) * g * t^2
We are interested in finding the vertical displacement (y). However, we don't have the time (t) directly. To find the time it takes for the ball to reach the catcher, we can use the horizontal distance and the horizontal velocity of the ball:
d = v0x * t
where:
d = horizontal distance
v0x = initial horizontal velocity
The initial horizontal velocity (v0x) is the same as the speed at which the ball is thrown, which is given as 46.2 m/s in this case.
Now, let's solve for the time (t) using the horizontal distance (d) and the initial horizontal velocity (v0x):
t = d / v0x
Substituting the values given in the question, we have:
t = 18.9 m / 46.2 m/s
t ≈ 0.409 s
Now that we have the time it takes for the ball to reach the catcher, we can substitute this value into the equation for vertical displacement to find out how much the ball will drop:
y = - (1/2) * g * t^2
Substituting the value of time (t) we found earlier:
y = - (1/2) * g * (0.409 s)^2
Using the approximate acceleration due to gravity, which is 9.8 m/s^2:
y ≈ -(1/2) * (9.8 m/s^2) * (0.409 s)^2
y ≈ -0.0805 m
Therefore, the ball will drop approximately 0.0805 meters by the time it reaches the catcher.