# Projectile Motion

1. Oct 13, 2009

### helpmee

1. The problem statement, all variables and given/known data
i have a problem from an in class lab where i have to find the initial velocity of the ball. given is the horizontal and vertical displacement of the projectile.
Dy(height)=100cm
Dx(range)=200m
0(theta)= 15degrees.
Find the initial velocity.

2. Relevant equations

3. The attempt at a solution
x-comp:
Vx1=Vo cos 60

y-comp:
Vy1=Vo sin 60

t=200/Vo cos 60

-100=Vo sin 60(200/Vo cos 60) + 1/2(-9.8)(200/Vo cos 60)^2
Vo= 2 numbers that make no sense

Last edited: Oct 13, 2009
2. Oct 13, 2009

### rock.freak667

Draw the initial velocity v0 at the angle 15 and then split that into vertical and horizontal components.

Then use your equations of motion.

3. Oct 13, 2009

### helpmee

hey
i havent learned equations of motion only kinematic equations

Last edited: Oct 13, 2009
4. Oct 13, 2009

### Redbelly98

Staff Emeritus
That's what he meant, they are the same thing. Use the kinematic equations.

5. Oct 13, 2009

### helpmee

i still dont understand
can u be specific on which ones to use?

6. Oct 13, 2009

### rock.freak667

the working in your original post looks fine, what values for v0 did you get? Also I don't understand how you have sin60 and cos60 when your angle is 15. Also the '-100' should just be 100

7. Oct 13, 2009

### helpmee

sorry that should be 15 not 60 and in class we learned that dy was also negative in this type of question

8. Oct 13, 2009

### rock.freak667

it was a lab, so I assume, you all weren't at the ground just shooting it at 15 degrees but did it ona a table at 100cm. In that case, you are right, it is -100.

So forming a quadratic in v0 what values of v0 did you get?

9. Oct 13, 2009

### helpmee

i got 2000m/s and 0.00125m/s. i think i might have done the calculations wrong though

10. Oct 13, 2009

### rock.freak667

yes check that back, because your equation shows you will get something like av02-b=0

11. Oct 13, 2009

### helpmee

can u help me on how to approach the math on this question?

12. Oct 14, 2009

### rock.freak667

ok so this what you have right?

$$-100=(V_o sin15)*(\frac{200}{V_o cos15}) + \frac{1}{2}(-9.8)(\frac{200}{V_o cos 15})^2$$

So in the first term there are we seeing anything canceling out? Can we simplify it if anything did cancel out?

in the second term, when you square out everything, can you find the number K such that the second term is K/Vo2 ?

13. Oct 14, 2009

### rl.bhat

In the problem D(y) is 100 cm. Is it correct?
Mainly point of projection is not given. If it is from the ground, at a given angle of projection and range, D(y) will be attained at different time depending on the initial velocity. So there cannot be a unique value of initial velocity.

14. Oct 15, 2009

### helpmee

Thanks for the help. i got it now