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Physics Angle Problem

  • #1

Homework Statement



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 ◦


Homework Equations



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

The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
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
i need to find the angle. i really do not know which equation to use...
 
  • #4
gneill
Mentor
20,793
2,773
What's special about the particular angle you want to find? What determines it?
 
  • #5
I dont know
 
  • #6
gneill
Mentor
20,793
2,773
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
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
gneill
Mentor
20,793
2,773
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
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
how far (delta y) the water balloon would go?
 
  • #10
gneill
Mentor
20,793
2,773
how far (delta y) the water balloon would go?
Yes. So what determines how high a projectile will go?
 
  • #11
the velocity ?
 
  • #12
gneill
Mentor
20,793
2,773
the velocity ?
Yes. What part of the velocity pertains to vertical motion?
 
  • #13
velocity along the y axis
 
  • #14
so would the equation be Vyf^2 = Vyi^2 + 2g(delta y)
 
  • #15
gneill
Mentor
20,793
2,773
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
would i have to isolate the formula?
 
  • #17
gneill
Mentor
20,793
2,773
would i have to isolate the formula?
You need to isolate Vyi from the formula.
 
  • #18
Vyi= sqrt of vyf^2 + 2g divided by y?
 
  • #19
gneill
Mentor
20,793
2,773
Vyi= sqrt of vyf^2 + 2g divided by y?
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)?
 

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