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.

 

[rwisz]

I would approach this problem by first identifying the given information and the unknown variables. In this case, we are given the initial velocity (38.2 m/s), the angle of launch (30 degrees), and the distance to be cleared (120m). The unknown variable is the height of the wall that the ball can clear.

Next, I would use the equations of projectile motion to solve for the unknown variable. The key equation to use here is the range equation, which relates the horizontal distance traveled (range) to the initial velocity, angle of launch, and acceleration due to gravity.

Using the range equation, we can rearrange it to solve for the height of the wall:

h = (d[x] * tanθ) - (g * d[x]^2) / (2 * v^2 * cos^2θ)

Where:
h = height of the wall
d[x] = horizontal distance (120m in this case)
θ = angle of launch (30 degrees)
g = acceleration due to gravity (9.8 m/s^2)
v = initial velocity (38.2 m/s)

Plugging in the values, we get:

h = (120 * tan30) - (9.8 * 120^2) / (2 * 38.2^2 * cos^2 30)
h = 35.4 m

Therefore, the tallest wall the player's ball can clear 120m away is 35.4 meters high. It is important to note that this calculation assumes ideal conditions and does not take into account air resistance or other external factors that may affect the trajectory of the ball.