Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 36.0 m/s and an initial vertical velocity of 21.0 m/s.
1) What is the initial speed of the cannonball?
2) What is the initial angle θ of the cannonball with respect to the ground?
3) What is the maximum height the cannonball goes above the ground?
4) How far from where it was shot will the cannonball land?
5) What is the speed of the cannonball 2.7 seconds after it was shot?
6) How high above the ground is the cannonball 2.7 seconds after it is shot?

Answer 1

1) Use the Pythagorean theorem.
Initial Speed = Vo = sqrt [Vx^2 + Vy^2]
2) Launch Angle = arctan(Vyo/Vx)
3) g*Hmax = Vyo^2/2
where Vyo is the initial vertical velocity component.
4) Range = 2*(Vo^2/g)*sinAcosA
5) Use the Pythagorean theorem again, but with the changed value of Vy.
Vy = Vyo - g*t
6) Height = Vyo*t - (g/2)*t^2