Projectile motion cannon shell problem

AI Thread Summary
A coastal defense cannon fires a shell horizontally from a 50.0m high cliff at a velocity of 60.0m/s. The vertical motion is calculated using the formula for free fall, resulting in a time of 3.2 seconds for the shell to hit the water. The horizontal distance is then calculated using the formula for distance, yielding a range of 192 meters. The calculations confirm the use of appropriate physics formulas for projectile motion. The final range of the shell's trajectory is therefore 192m.
Zugox
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Homework Statement


A coastal defence cannon fires a shell horizontally from the top of a 50.0m high cliff directed out to sea with a velocity of 60.0m/s.
Determine the range of the shell's trajectory.

Homework Equations


delta x = ut (horizontally)
delta y = ut + 1/2 at^2

The Attempt at a Solution


working vertically
50 = 0t + (1/2 * 9.8 * t^2)
t = 3.2

working horizontally
delta x = ut
delta x = 60 * 3.2
delta x = 192
therefore the range is 192m

I'm not too great with physics although really willing to learn and really interested in it :)
Can someone just confirm I'm right or put me on the right track
 
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I haven't done the math but your work seems right, I'm studying this area in Physics class too :D
 
MysticDude said:
I haven't done the math but your work seems right, I'm studying this area in Physics class too :D

I'm 99% sure on the math, was just worried about the formulas I used:)
 
Well, since we are firing horizontally, \theta = 0 meaning y-y_0 = \frac{-1}{2}gt^2 and you have that. Then you can just plug in for d=vt and you should get 192m.
 
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