The angle and distance of a projection - Grade 11 physics

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Homework Help Overview

The problem involves a water balloon being projected from a cannon at a specific speed and angle, with the goal of determining whether it will hit a target located a certain distance away. The subject area is projectile motion in physics.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for the horizontal distance the balloon travels and question whether the approach to adjusting the cannon's speed or angle is optimal for hitting the target.

Discussion Status

Some participants have provided feedback on the calculations and are exploring alternative parameters that could be adjusted to improve accuracy. There is an ongoing dialogue about the practicality of changing the angle versus the speed of the cannon.

Contextual Notes

The original poster is required to justify their calculations and suggestions, which may influence the direction of the discussion. There is a lack of explicit consensus on the best parameter to adjust for achieving the target hit.

CAT 2
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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 = v1To make the balloon hit the target the speed needs to be decreased to 26.5m/s.Is this done correctly?
 
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Looks good.

What other parameter might you vary to change the range of the water canon? Which parameter might be easier to vary accurately (on a real water canon)?
 
The angle would probably have been a better choice, but it was easier to adjust the speed. They don't exactly specify which one to change, (speed or angle). Do you think it will matter?
 
CAT 2 said:
The angle would probably have been a better choice, but it was easier to adjust the speed. They don't exactly specify which one to change, (speed or angle). Do you think it will matter?
Probably not in a purely technical sense. But I would think that practically speaking, a canon would have targeting swivels (altitude and azimuth) built into its design, and simple to adjust. Changing the "charge" of a shot might be more cumbersome.
 
Yeah, that's true. Thanks for your help.
 

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