Determine how high the ball is (from the ground)

  • #1
My professor decided to throw us a curveball to try and solve the following problem:

A third baseman makes a throw to first base 38m away. The ball leaves his hand with a speed of 36m/s at a height of 1.7 meters from the ground and making an angle of 14 degrees with the horizontal.

a) Determine how high the ball is (from the ground) when it gets to the first base.
b) Is the throw successful

No equations given.

My shot at the problem

a) It would be a parabolic function because of its projectile nature. Thus the equation using H,R is y=a(x-1/2R)^2 + H. where a = -g/2(V0)cos^2(theta).

First i solved R:
R=Vo^2(sin(2theta))/g
=(36m/s)^2(sin(28))/9.8m/s2
= 62.085m

Then i solved for H
I used h as H-1.7 due to the fact that the curve only applies to that area over the basemans arm to the next point where H = 1.7.

h=Vo^2(sin^2(theta))/2g
h=(36m/s)^2(sin^2(14))/19.6m/s2
h=3.87 + 1.7= H= 5.57

So here's the parabolic function now:

y= (-(9.8m/s2)/2(36m/s)^2(cos^2(14deg)))(x-.5(62.085m))+ 5.57m
y= (-.004 no units)(x-31.0425m)+5.57m
f(38 [length from 3rd to 1st baseman]) = (-.004*6.9575)+5.57 {m}
f(28) = 5.54217 m above the base.

So therefore i got the answer to b) being that no, the catch wasn't made because the ball is three times higher than a normal human height (1.8m), thus the catch was not made.

Hopefully i can get help with this... i appreciate it.
 

Answers and Replies

  • #2
russ_watters
Mentor
20,577
7,240
Checking by a different method, by calculating the horizontal and vertical velocity components: 36 sin (14) & 36 cos (14) , we can see from the results that the ball gets to first base in just over a second, just past the top of it's arc. That's only halfway toward calculating the final height, but certainly it would not be catchable.
 

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