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?
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
Once you know time, you plug that in to find the horizontal distance.
Δdx=Vix X Δt
This is the answer because I did not include the 38.0m. How can i do that? The correct answer is 736m.