Projectile Motion: Speed Difference between Horizontal and Vertical Throws

[J.live]

Homework Statement

Two baseballs are thrown off the top of a building that is 7.32 m high. Both are thrown with initial speed of 58.7 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the two balls when they touch the ground? (Neglect air resistance.)

vball 1 - vball 2 = ?

Homework Equations

Ei = Ef

The Attempt at a Solution

58.7 mph -> 26.08 m/s ? (not sure)

I used

Ei= Ef

Ball 1 ( x - direction)

Ki (0) + Ui (mgh)= 1/2mv^2+ Uf(0)

gh= 1/2 V^2

V= 11.97m/s ( same velocity for Ball 2 (y-direction))

I just don't know how to find the difference. If someone can explain it in steps. Thanks

 

[J.live]

bump :/

 

[SiYuan]

58.7 mph = 26.241m/s
This is not an energy problem per se, more of a kinematic one.


Since air resistance is negligible, you can assume that the horizontal velocity of the ball thrown horizontal to be constant,

Does that help now?

 

[J.live]

so the velocity for horizontal is going to stay 26.241 m/s, right?

Then I just simply subtract the two velocities?

 

[SiYuan]

Yep. Since the question is asking for the difference in speed, as Ball 1 falls to the ground it gains vertical velocity as well (how else would it fall?), and Ball 2 will fall to the ground, gaining vertical velocity.

 

[J.live]

Well I subtracted 11.97 (ball1) - 26.241 (ball2) = -14.271. It came out to be wrong

The answer is -9.37 m/s :/

 

[SiYuan]

No no, that's not quite how you do it.

Let's keep the horizontal velocity of both balls out of the picture for now.

The vertical velocity of Ball 1 will increase from 0 to X as it falls to the ground,

whilst the vertical velocity of Ball 2 will increase from 26.24(thrown to the floor) to (26.24 + Y) as it falls to the ground. Calculating it is simply, v^2 = u^2 + 2AD where A is acceleration due to free fall and D is the height of the building.

After which, we look at horizontal velocity.

Since the horiz velocity of ball 2 is 0, and will likely stay that way.

The horiz of Ball 1 is constant, which is 26.241.

Hence, just using Pythagoras' on Ball 1, you can find the speed for Ball 1.

The speed of Ball 2 would be the vertical velocity only.

 

[J.live]

what the? We have to find the vertical speed of ball 1? Why do we have to do that?
so using Pythagorean

(7.3)^2 + (26.241)^2 = c^2 ?

I am not getting this.

 

[SiYuan]

Your vertical velocity (7.3) for Ball 1 is wrong.

 

[J.live]

No, I am not to able to understand what we are looking for here? Why do we need to find the vertical velocity for ball 1 in the first place? If so how do we do it? using Pythagorean ?

 

[SiYuan]

To find the vertical velocity for Ball 1, you engage the same method as Ball 2, where v^2 = u^2 + 2AD,

Since initial vertical velocity of Ball 1 = 0, hence V^2 = 2AD, V = 11.98m/s

The final velocity of Ball 1 would be C^2 = 11.98^2 + 26.241^2

 

[J.live]

oh, so that's how we find the overall velocity for Ball 1 , right?

Then for Ball 2? We have to find the overall velocity for that too?

How would we do that ?

 

[SiYuan]

Yep, as I mentioned earlier, the velocity for Ball 2 would be the initial velocity + the additional vertical velocity, which is found by V^2 = U^2 + 2AD as well

 

[J.live]

I got 11.98 m/s which is the final velocity for ball 2 as well?

I don't understand how can we simply add 26.41 to 11.98 for ball 2? Isn't it two dimensional like Ball 1?

 

[SiYuan]

Yes, it is 2 dimensional, but Ball 2 experiences only vertical velocity, since it is thrown straight down.

 

[J.live]

makes sense. So in that case we can add the two velocities without using Pythagorean ?

 

[SiYuan]

If you did the kinematics method (V^2 = U^2 + 2AD), you should get the final velocity which is after addition already.

 

[J.live]

For Ball 1 11.98^2 + 26.241^2= c^2 ---> 28.84

For Ball 2 using the same method V^2 = U^2 + 2AD
V^2= 26.241^2 + 2 (9.8)(7.32)
V = 28.84 m/s

I still didn't get the right answer.

What am I doing wrong?

Sorry for bothering you time and again.

 

[SiYuan]

What is the answer?

 

[J.live]

Answer given is -9.37 m/s

 

[SiYuan]

Have you missed out anything from the question?

 

[J.live]

No, I have not missed out since I copy pasted it.

What the professor did was use the Kinetic energy equation

Ball 1
Eiy = Efy

Ki + Vi = Kf + Vf

0+ mgh = 1/2 mvf^2+0

Vfy= 11.98 m/s
================
Ball 2

Eiy= Efy

0+mgh= mvf^2 + 0

Vfy= 11.98 m/s
============================
Then he used Pythagorean Theorem for Ball 1

11.98^2 + 26.241^2= C^2

But for Ball 2 He simply added Viy(26.241)+ VfY(11.98)

Which does come out to be -9.38 m/s :/

 

[SiYuan]

No, I have not missed out since I copy pasted it.

What the professor did was use the Kinetic energy equation

Ball 1
Eiy = Efy

Ki + Vi = Kf + Vf

0+ mgh = 1/2 mvf^2+0

Vfy= 11.98 m/s
================
Ball 2

Eiy= Efy

0+mgh= mvf^2 + 0

Vfy= 11.98 m/s
============================
Then he used Pythagorean Theorem for Ball 1

11.98^2 + 26.241^2= C^2

But for Ball 2 He simply added Viy(26.241)+ VfY(11.98)

Which does come out to be -9.38 m/s :/

There is something faulty about this.

The initial KE of Ball 2 should not be 0, but instead 1/2 (m)(26.241)^2.

And after calculation, the final velocity still gives me the same answer I had during kinematics.

Your professor has made the mistake that (Vf + Vi)^2 = Vf^2 + Vi^2

 

[J.live]

This is strange. Anyways, thanks for your help.