# A ball being hit, find the energies and maximum height

Tags:
1. May 31, 2016

### Charlene

1. The problem statement, all variables and given/known data
The longest homerun hit by Miguel Cabrera in the 2012 season occurred at Comerica Park and had an initial y-component of velocity of 19.8 m/s and an initial x-component of velocity given as m/s, where is some value left undetermined.
Assume the ball left the bat 1.22 m off of the ground. (Ignore the effects of air resistance.) Give the following:
a) the initial kinetic energy
b) the initial gravitational potential energy
c) the kinetic energy at the maximum height
d) the gravitational potential energy at the maximum height
e) the maximum height above the ground (you must find this using energy calculations)

2. Relevant equations
a) KI= 1/2 m1v1^2+ 1/2 m2v2^2
b) GPE= mgh
c)When the ball is at maximum height the velocity is 0 therefore i think the kinetic energy is also 0?
d) When the ball is at maximum height the velocity is 0 therefore i think the gravitational potential energy is 0..?
e)hmax=v^2(sin^2(θ))/2g

3. The attempt at a solution
a)KI=(1/2 m1(19.8m/s)^2) + (1/2 m2 (v m/s)^2)
=9.9m^2/s^2 (m1) +(m2 v m^2/s^2)/2
I feel like im just missing a lot of information and don't understand how to find the mass of the ball..
b)GPE initial= mgh
(m)(9.8 m/s^2)(1.22m)
Once again i don't have mass so i'm confused where to go from here
c) 0
d) 0
e) i don't know if the formula i've shown is using energy calculations or not, and if it is i don't understand how to find θ and what velocity to use.

2. May 31, 2016

### BvU

Hi Charlene,

I think you are supposed to use symbols for the unknowns. So the answers are experssions instead of numbers (with dimensions)

a) there is only one ball, so only one $m$
c) is it really hanging still at that point ?
d) conflicts with the expression in b) (unless highest point is 0 m high ?)
e) re-do after the others are fixed.

3. May 31, 2016

### Charlene

So for part a) would it make sense for my answer to be 1/2m(392.04 m^2/s^2 +v^2 m^2/s^2)?

and for c) i do feel like the ball is sitting still at the maximum height because the ball slows down to a stop before falling back toward the ground.

d) so i need to find maximum height BEFORE i can actually solve for d?

e) h=V(vert)^2 sin(θ)/2g
h=392.04 m^2/s^2 sin(θ) / 2(9.8 m/s^2)

Last edited: May 31, 2016
4. May 31, 2016

### BvU

c) What makes it stop having a horizontal speed v ?
d) no. Think symbols. What is conserved ?

5. May 31, 2016

### Charlene

yes your correct, it doesn't stop having a horizontal speed i agree. but i thought at the maximum height i'd be looking at the vertical speed?

as for d) my initial guess would be to find a max height to plug into the equation for h, but since i'm not supposed to do that, I would think that maximum gravitational height would be equal to the maximum height because mg can be constants and would cancel out?

6. May 31, 2016

### BvU

The exercise asks for

7. May 31, 2016

### Charlene

so the kinetic energy at maximum height is
KE=mv^2
because i'm unaware of the mass and im given v as my velocity for the horizontal

i now see why its not 0, it would only be 0 if it was thrown straight in the air, correct?

as for the gravitational potential energy,

GPE=mgh
=m(9.8m/s^2)h

(19.8 m/s)^2=0 + 2 (9.8 m/s^2)d
d=20 m

therefore GPE= m(9.8)(20)
=m(196 m^2/s^2)
is this now the correct approach

Last edited: May 31, 2016
8. May 31, 2016

### BvU

Actually it is $\ {1\over 2}mv^2 \ \$ with $v\$ the initial horizontal speed.
Correct, then the horizontal speed is 0.

d)
GPE = mgh = m(9.8m/s^2)h $\qquad$ yes. But there is more to do here.
h was a length, so this doesn't match dimensionally. (what is h anyway ?). Also you don't say what ad is (but I can guess -- looks like one of the kinetic equations.). Why introduce different names for these variables ?

So it looks as if Vf is the initial vertical speed, and Vi is the vertical speed at the highest point. Then ad is actually $g\Delta h$, the gravitational acceleration times the maximum vertical travel. I don't see the given 1.22 m at the start ?

A decent way to make this recognizable is to write this as an energy balance:$$mgh_0 + {1\over 2}mv_{\rm 0,\, y}^2 = mg h_{\rm max} + {1\over 2}mv_{\rm f,\, y}^2 \Rightarrow mgh_{\rm max} = mgh_0 + {1\over 2} mv_{\rm 0,\, y}^2$$ (after all d asks for the gpe).

e)
Well, you have the gpe from d), so divide by gh to not get the 20 m you already found, but ...

9. Jun 1, 2016

### Charlene

Thank you so much, i understand it now! So using the formula you provided

hmax=ho+(mv^2)/2mg
=1.22 m + (19.8 m/s)^2/(2(9.8m/s^2))
Max height =21.2 m

And then plugging this into the GPE at max height would be
m(9.8m/s^2)(21.2m)
=(208. m^2/s^2) m

10. Jun 1, 2016

### BvU

Good! makes my day .