# Projectile motion- only distance is known

1. Feb 13, 2010

### sean-820

1. The problem statement, all variables and given/known data

A baseball, thrown from shortstop position to first base travels 32m horizontally, rises 3mm and falls 3m. Find the velocity of the ball.

2. Relevant equations

d=v(t)+0.5(a)(t)^2

3. The attempt at a solution

So i know horizontal motion will be a constant velocity and vertical will have acceleration
(-9.8m/s/s)

For the equation i know d=2.997m, a=-9.8m/s/s but there are still two unknowns. I was thinking of substitution or elimination to get rid of a variable, with anything i sub in has a variable so it gets no where.

If there was one more given it would be easy, but i cant get anywhere with just displacement other then the time it takes just to drop it (which should be the time when its dropped 3mm to the ground but i dont know that smaller distance.

Gan somebody help me with what my next step should be?

2. Feb 13, 2010

### kuruman

Your problem is not that you don't have enough givens' it is that you have not listed enough relevant equations. What other equations do you know about projectile motion?

Are you sure that the ball rises 3 mm and falls 3 m? Try to picture this. If the ball falls an overall distance of 2.997 m, from what height was it launched? How tall is a shortstop?

3. Feb 13, 2010

### sean-820

Would the shortstop not be 3m-3mm tall?

Other formulas:
v=d/t
v2^2=v1^2+2a(d)
quadratic formula- (-b+- sqrt(b^2-4ac))/2a
but i dont have enough info to sub in.
I looked in my notes and textbook and the formula i gave is the one we are using. i didnt see any other formulas other then kinematics formulas like the ones i gave above.

I dont even know the angle it is launched at other then the projectiles peak

The exact question is A baeball thrown from shortstopposition to first base, travells horizontally 32m, rises 3mm and falls 3m. Find the initial velocity. 3mm does sound like it could be a typo as its a small distance, but its whats in the question.

4. Feb 13, 2010

### kuruman

Probably not. Three meters is about ten feet. Considering that shortstops release the ball from about shoulder height, this particular shortstop would have to be about twelve feet tall. The tallest person in the world, ever, is less than nine feet.

Be that as it may, since the silly numbers you are given are what you have to work with, you should be able to find the vertical component of the initial velocity using one of the equations that you have listed so far. How do you think you can do that?

5. Feb 13, 2010

### sean-820

would mgh=0.5mv^2 work too im assuming. Mass can cancel out. and i know gravity and height so the only variable would be vertical component. Now that you point it out it seems much easier as i was trying to find the horizontal component first.

The question isn't worded the best as it could also mean the ball hits the ground after 32m if the ball travels 32m per 3mm rise then in addition to that it falls

thanks for the help btw.

6. Feb 13, 2010

### kuruman

That would work. I trust you can finish the problem now.

7. Jul 17, 2011

### 5.98e24

I hope it's okay to post in this thread even though it's 1.5 years old, but which equation is kuruman referring to, that you can use in order to determine the vertical component of the velocity? Is it the quadratic formula?

8. Jul 17, 2011

### amy andrews

Use the same equations as for horizontal velocity, with the acceleration equal to g.

9. Jul 17, 2011

### 5.98e24

But doesn't horizontal velocity have no acceleration, since the speed is constant? Which horizontal velocity equation has acceleration in it?

10. Jul 17, 2011

### amy andrews

Basic constant acceleration equations:
v=v_0+at
x=x_0+v_0t+(1/2)at^2
v^2=(v_0)^2+2aΔx

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