Calculate Horizontal Ball Drop at 41.3m/s Velocity

[pookisantoki]

A Major league pitcher can throw a ball in excess of 41.3m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0m away from the point of release?
So Ay=-9.8
X=17m
t=?
Vox=0
V=41.3
y=(Vy^2-Voy^2)2g
I thoguht i would use this formula but wasnt sure how to figure Voy out since Vy=0...
Please help!

 

[Redbelly98]

X=17m
t=?
Vox=0

Vox, the initial horizontal speed, is not 0, it is ____?
Also, ax = ____?

Can you fill in the blanks, and then use this information to find t?

 

[tomcue]

To calculate the horizontal ball drop at a velocity of 41.3m/s, we can use the formula d = v*t, where d is the distance traveled, v is the velocity, and t is the time. In this case, the distance traveled is 17m and the velocity is 41.3m/s. Therefore, we can rearrange the formula to solve for time:

t = d/v = 17m/41.3m/s = 0.411 seconds

Now, to calculate the vertical drop, we can use the formula y = Voy*t + 0.5*Ay*t^2, where y is the vertical distance, Voy is the initial vertical velocity, Ay is the acceleration due to gravity (-9.8m/s^2), and t is the time calculated above. We know that Voy = 0, so the formula becomes y = 0 + 0.5*(-9.8m/s^2)*(0.411s)^2 = -0.83m.

Therefore, the ball will drop approximately 0.83m when it reaches the catcher who is 17m away from the point of release. This is assuming that there is no air resistance and the ball is thrown in a straight line.