Projectile motion cannon shell problem

[Zugox]

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

 

[MysticDude]

I haven't done the math but your work seems right, I'm studying this area in Physics class too :D

 

[Zugox]

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:)

 

[MysticDude]

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.

 

[rwisz]

if I'm wrong?

Your solution is correct. The only thing to note is that the range of the shell's trajectory would be the distance from where it was fired to where it lands, which would be a bit more than 192m in this case. This is because the shell would also travel a distance of 50m vertically before landing, so the actual range would be 192m + 50m = 242m.