# Projectile Motion question

1. Oct 28, 2003

### pulau_tiga

can you please help me with this question. I really am stumped, on how to get the initial velocity.

A ball player hits a home run and the baseball just clears a wall 21.0 m high located 130 m from home plate. (assume max height of the ball is 21.0 m) The ball is hit at an angel of 35.0* to the horizontal at a height of 1.0 m.
a) What is the initial velocity of the ball?
b) how much time does it take the ball to reach the wall?
c) Find the veolicty components and the velocity of the ball when it reaches the wall.

Thanks

2. Oct 28, 2003

### enigma

Staff Emeritus
Hi pulau_tiga!

Welcome to the forums!

Can you post what you've tried so far, even if it's just looking up the relevant formulas?

You won't learn it if we do the problem for you.

3. Oct 29, 2003

### pulau_tiga

Okay.
I know that a=-9.8 m/s; and d= 20.m
but I don't know if Vf = 0... I think it might, but I'm not sure.
If it does = 0 then I think I can do...
Vf2 = Vi2 +2ad
0 = Vi2 +2ad
-2ad = Vi2
-2(-9.8)(20.) = Vi2
390 m2/s2 = Vi2
Vi=20. m/s
but I don't know if Vf = 0 or not.
my answer seems reasonable, so that seems possible to me.

4. Oct 30, 2003

### HallsofIvy

Staff Emeritus
No, it is not necessary that vf be 0. If we assume that the ball is at its highest point when it passes the wall that would be true but you can't assume that. By the way, did you use the fact that the initial angle is 35 degrees?

Unfortunately, I don't know what formulas you have available to use.

Here's how I would do the problem.

Calling the initial speed v0, we know that the initial "velocity vector" was (v0 cos(35), v0 sin(35))- that is, that the initial horizontal speed was v0 cos(35) and the initial vertical speed was v0 sin(35). We also know that the horizontal acceleration is 0 and that the vertical acceleration is -9.81 m/s^2.

From that, we can get that the horizontal speed at any time, t seconds, is v0 cos(35) (no acceleration so it doesn't change) and that the vertical speed at time t is vo sin(35)- 9.81t.

From that, we can get that the horizontal position at any time, t seconds, is vo cos(35)t and that the vertical position at time t is
1+ vo sin(35)t- 4.9 t^2 (That "1" is the initial height).

Since the wall is 130 m horizontally and 21 m high, and the ball "just cleared" the fence, at that time we must have
v0 cos(35)t= 130 and 1+ v0 sin(35)t- 4.9t^2= 21.

You can solve those two equations for v0 and t.

5. Sep 27, 2004

### Brando05

Hello, I am having trouble with this same exact question.
I read this topic fully and I do not get it.
The person gave two equations having two unkown variables, initial velocity and time in order to find time and initial velocity. How is this possible?
Can someone explain this in...eh...easier terms?
:uhh:

Thanks,
Brandon

6. Sep 27, 2004

### Brando05

Also...where did the 4.9 come from in that one equation...

7. Sep 28, 2004

### Brando05

OK, so 4.9 is half of 9.81...

8. Sep 28, 2004

### cdhotfire

hehehe, that was a dozy 9.8/2=4.9. Whew. :tongue2:
jk

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