- #1
JackandJones
- 16
- 0
The Question:
A water balloon is fired 34 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) is painted on the asphalt 42m away from the water cannon.
a) Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land.
b) make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.
For part a)
I found total time to be 2.1 s
Water balloon lands at 67.8 m (therefore it does not hit the target that is 42 m away)
For part b)
I understand that when angles are higher than 45 degrees, horizontal displacement decreases and the max vertical height increases.
I don't know how to find the new launch angle that will allow the balloon to hit the target at 42 m. Any help is appreciated!
A water balloon is fired 34 m/s from a water cannon, which is aimed at an angle of 18° above the ground. The centre of the cannon's target (which has a radius of 1.0m) is painted on the asphalt 42m away from the water cannon.
a) Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land.
b) make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.
For part a)
I found total time to be 2.1 s
Water balloon lands at 67.8 m (therefore it does not hit the target that is 42 m away)
For part b)
I understand that when angles are higher than 45 degrees, horizontal displacement decreases and the max vertical height increases.
I don't know how to find the new launch angle that will allow the balloon to hit the target at 42 m. Any help is appreciated!