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## Homework Statement

A cannonball is launched at 90.0m/s at an angle of 60.0° above the horizontal from the top of a tower that is 38.0m high. How far away from the tower does it land if the ground is level?

## Homework Equations

Vix=90.0m/s X Cos60

Viy=90.0m/x X Sin60

## The Attempt at a Solution

I used the formula to find the distance from the tower, but i did not include the initial 38.0m that it was launched from. I do not know how this fits into the equation.

x-comp: Vix=90.0m/s X Cos60= 45m/s

y-comp: Viy=90.0m/s X Sin60=77.94228634m/s

Δdy=Viy×Δt+1/2a(t)^2

0=77.94228634m/s(Δt)+1/2(-9.81m/s^2)(Δt^2)

Δt=(77.94228634m/s)/(4.905m/s^2)

t=15.89037438s

Once you know time, you plug that in to find the horizontal distance.

Δdx=Vix X Δt

Δdx=(45m/s)(15.89037438s)

Δdx= 715.0668471m

This is the answer because I did not include the 38.0m. How can i do that? The correct answer is 736m.