# Grade 11 Physics Projectile Motion Question

1. Apr 23, 2013

### mmChocolate

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Vix=90.0m/s X Cos60
Viy=90.0m/x X Sin60

3. 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.

2. Apr 23, 2013

### Yosty22

The Initial height is in the position equation, not the velocity equation. Are you familiar with that equation?

3. Apr 23, 2013

### Yosty22

To clear that up, use the y-component of the position equation to solve for time. That way, it takes into account the initial height of the ball. Then, using the components of velocity given and the known acceleration, you should be able to figure out the rest fairly easily.

4. Apr 23, 2013

### mmChocolate

Thanks! That helped a lot.

5. Apr 23, 2013

### Yosty22

Just remember the equations for the future. All of the equations are fairly simple to work with. Therefore, if your answer doesn't really make sense with one equation because of either a left-out variable or something else, there are other equations to try if you have the variables.