Projectile Motion Launched at an Angle

1. Oct 3, 2011

phizics09

1. The problem statement, all variables and given/known data
A golf ball is launched from the roof of a school with a velocity of 20 ms at an angle of 30 deg above the horizontal.

2. Relevant equations
If the roof is 40m above the ground, calculate
a) the ball's time of flight
b)the ball's horizontal displacement

3. The attempt at a solution
So for the first question, the example in the book only uses one equation to solve for t, D=v1(t)+1/2a(t)^2,but I thought since the ball goes up and comes down, there should be two separate parts to the solution?

2. Oct 3, 2011

1MileCrash

No, it's a single parabola. However, we should still break the initial vector they give you into components to separately model x and y positions.

Once you've done that, you can set the equation modeling y position to 0, then find the value of t that makes it true, and you have time of flight.

Given that time for t, you can plug that into the x position equation and you have horizontal displacement.

Do you know what to do with the vector they've given you?

3. Oct 3, 2011

phizics09

Yeah, I get it now. So if it wasn't a projectile problem, I would still use the same D for the entire question? For example, if a kinematics problem has a ball thrown upwards at the edge of a the top of a building with an initial vertical velocity, and the ball lands on the ground below, and it asks you to find the time it takes to land on the ground, you would just use the height of the building in the equation D=v1t+1/2a(t)^2 to find t?

4. Oct 3, 2011

1MileCrash

Yes, your equation for the y position of the projectile will only depend on the initial y velocity and the initial position, but nothing to do with the x position/velocity.

You can always break it into components this way. Just remember to do it correctly. Launching it at 20 m/s at a 30 degree angle is the same as launching it at 20sin30 m/s vertically.

5. Oct 4, 2011

Thanks!