- #1

CAT 2

- 44

- 1

## Homework Statement

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.

## Homework Equations

This equation is used for finding horizontal distance.

Dh = v1^2 Sin 2(angle)/ a

## The Attempt at a Solution

. https://www.physicsforums.com/file:///C:/Users/User/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png v1 = 34 m/s

angle = 18°

a = 9.8 m/s^2 [down]

Find the horizontal projected distance of the projected balloon

Dh = v1^2 Sin 2(angle)/ a

Dh = (34 m/s)^2 (Sin2(18°)/ 9.8 m/s^2

Dh = 69.3 meters

Since the distance to the target is 42 meters and the water balloon traveled 69.3 meters, the balloon did not hit the target.

To make it hit the target, the speed needs to be decreased.

Dh = v1^2 Sin 2(angle)/ a

42m = v1^2 (Sin2(18°)/ 9.8 m/s^2

42m = v1^2 (0.05998)

√700 = v1

26.5 m/s = v1

To make the balloon hit the target the speed needs to be decreased to 26.5m/s.

Is this done correctly?