# Physics Angle Problem

1. Oct 13, 2013

### mithilsheth

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

Some enterprising physics students working
on a catapult decide to have a water balloon
ﬁght in the school hallway. The ceiling is of
height 3.3 m, and the balloons are launched
at a velocity of 9.7 m/s. The acceleration of gravity is 9.8 m/s2 .At what angle must they be launched to just graze the ceiling?Answer in units of ◦

2. Relevant equations

Im not quiet sure which equation to use, all of projectile motion is really tough for me.

3. The attempt at a solution

2. Oct 13, 2013

### Staff: Mentor

Sorry, you must make some effort.

Draw a picture. Describe which features of the trajectory are important to answering the question (what do you need to find out)? What equations pertain to those features of the trajectory?

3. Oct 13, 2013

### mithilsheth

i need to find the angle. i really do not know which equation to use...

4. Oct 13, 2013

### Staff: Mentor

What's special about the particular angle you want to find? What determines it?

5. Oct 13, 2013

### mithilsheth

I dont know

6. Oct 13, 2013

### Staff: Mentor

Well, if you were one of the students firing the projectiles, how would you determine if the shot met the criteria of the problem? What condition is placed on the desired angle by the problem statement?

7. Oct 13, 2013

### mithilsheth

the water balloon should just graze the ceiling which is 3.3 meters tall and is going at a initial velocity at 9.7 m/s

8. Oct 13, 2013

### Staff: Mentor

Yes! So what part of the trajectory corresponds to "just grazing the ceiling"? What would you need to find out about a projectile launched at some angle with some speed to know if it would satisfy the requirement?

9. Oct 13, 2013

### mithilsheth

how far (delta y) the water balloon would go?

10. Oct 13, 2013

### Staff: Mentor

Yes. So what determines how high a projectile will go?

11. Oct 13, 2013

### mithilsheth

the velocity ?

12. Oct 13, 2013

### Staff: Mentor

Yes. What part of the velocity pertains to vertical motion?

13. Oct 13, 2013

### mithilsheth

velocity along the y axis

14. Oct 13, 2013

### mithilsheth

so would the equation be Vyf^2 = Vyi^2 + 2g(delta y)

15. Oct 13, 2013

### Staff: Mentor

That would be a good choice. Be careful with the directions assigned to the motion and acceleration.

Can you determine from that equation the Vyi required to just reach your given height?

16. Oct 13, 2013

### mithilsheth

would i have to isolate the formula?

17. Oct 13, 2013

### Staff: Mentor

You need to isolate Vyi from the formula.

18. Oct 13, 2013

### mithilsheth

Vyi= sqrt of vyf^2 + 2g divided by y?

19. Oct 13, 2013

### Staff: Mentor

That doesn't look quite right. But it's hard to tell what you intended without parentheses to fix the order of operations. What's divided by y, The 2g? What's inside the square root?

Use parentheses to gather together the terms to fix the interpretation of the expressions.

Can you show step-by-step the isolation of Vyi? You might also consider taking care of Vyf early on. What's the value of Vyf (when the projectile reaches its maximum height)?