Solving for the Vertical Distance of a 101 mph Pitch

[toothpick09]

Homework Statement

One of the fastest recorded pitches in major-league baseball, thrown by Billy Wagner in 2003, was clocked at 101.0 mi/h (Fig. P3.22). If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically (in feet) by the time it reached home plate, 60.5 ft away?

Homework Equations

t impact = 2 * (vo/g) sin theta

The Attempt at a Solution

2 * (9.21404m (60.5ft)/9.8m/s2) sin(90) = 1.88042 m

1.88042 m * 3.281 ft = 6.16965 ft

what am i doing wrong?

 

[hage567]

For one thing, it looks like you are mixing ft and meters. What is the 9.21404 m supposed to be?

Your formula is for "t impact", but you stated the answer in meters?? I think you have confused things. I don't think your formula is correct, either. Can you show how you got it?

 

[m2003]

I would approach this problem by first converting all the given units to a consistent system. In this case, since the velocity is given in miles per hour and the distance in feet, I would convert the velocity to feet per second by multiplying it by 1.47 (1 mile = 5280 feet, 1 hour = 3600 seconds). This gives us a velocity of 148.47 ft/s.

Next, I would use the equation for distance traveled under constant acceleration, d = v0t + 1/2at^2, where v0 is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2 or 32.2 ft/s^2), and t is the time. In this case, we are solving for the vertical distance, so we can set the initial velocity to 0 since the ball is thrown horizontally.

Plugging in our values, we get:

60.5 ft = 0 + 1/2(-32.2 ft/s^2)t^2

Solving for t, we get t = 1.53 seconds.

Now, we can use the equation for time of impact, t_impact = 2 * (v0/g) * sin(theta), where v0 is the initial velocity and theta is the angle of launch (90 degrees in this case). Plugging in our values, we get:

t_impact = 2 * (148.47 ft/s / 32.2 ft/s^2) * sin(90) = 9.22 seconds.

Finally, we can use the equation for distance traveled under constant acceleration again, this time solving for the vertical distance:

d = 0 + 1/2(-32.2 ft/s^2)*(9.22 s)^2 = 149.5 ft

Therefore, the ball would fall approximately 149.5 feet vertically by the time it reaches home plate, assuming it is thrown horizontally at a speed of 101 mph.