Solving a Simple Height Problem for a Home Run in Baseball | 120m Distance

[mohabitar]

A baseball player wants to hit a home run over the far wall of a stadium. He hits the ball 1 meter above the ground so that its speed is 38.2 m/s and such that it makes an angle of 30 degrees with respect to the horizontal. What is the tallest wall the players ball can clear 120m away?

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So I tried breaking down the velocity into its vertical and horizontal components, to get V[x]=33.08 and V[y]=19.1 m/s.

Then I used basic equations to find whatever values I could find, without any particular goal. So if d[x]=120 and V[x] 33.08, and final velocity is 0, then I can use v^2=u^2+2as, to get an acceleration of 4.55 m/s.

But really after this I had no idea where to head. Am I on the right path?

 

[cepheid]

The goal here is to compute the vertical height of the ball when it is a horizontal distance of 120 m away from the launch point. This vertical height tells you the tallest wall that can be cleared. Does that make sense?

 

[mohabitar]

Well you I knew that part I just meant I was solving equations with no particular reason. How should I set this up?

 

[cepheid]

You know equations for the height y vs time and the horizontal distance x vs time (in terms of the given parameters including launch speed, launch angle, and initial height).

If you can figure out the time at which x = 120 m, it stands to reason that you can then figure out the vertical height at that time.

Edit: the method you tried is incorrect since the final velocity (at max height) is not zero in the x-direction, so you have misapplied the formula you used. Furthermore, there is no acceleration in the x-direction, which means the formula you used does not apply at all.