Calculating Projectile Motion for Water Balloon Launch

[yaho8888]

This problem is about shooting water Balloons!
If a balloon is fire from an 45 degress angle and it's Range is about 52m and It took about 2.9sec to get to that range. How can I find the initial velocity, the Velocity of vertical, the Velocity of Horziontal, and the Height the Ballon goes?

Show the work please!

THanks!

 

[yaho8888]

I have found out the horziontal veocity already is 17.93m/s some one please help!

 

[SGT]

The horizontal velocity is $$V_0 cos(45)$$

 

[HallsofIvy]

You are, I guess. assuming no air resistance. I'm not at all sure that's reasonable for water balloons, but okay. Since there is no horizontal force, the horizontal speed is constant and equal to 52m/2.9s. That's where you got 17.93 m/s. Now, as Sgt said, that' equal to Vcos(45) so you need to solve Vcos(45)= 17.93 for V. Actually, because a 45 degree right triangle is isosceles it should be easy to see that the vertical component of initial velocity is also 17.93 m/s.

Can you get the max. height from that?

 

[yaho8888]

NO i can't that is the only thing I found out!

 

[SGT]

NO i can't that is the only thing I found out!

Try to equate vertical velocity taking in account initial velocity and acceleration (gravity).
At what time will the velocity be zero? At that instant maximum height is achieved.

 

[yaho8888]

Thanks I got all of them!