On a 100 meter high cliff, a cannon fires a cannonball with an initial speed of 50 m/s at an angle of 53° to the horizontal. The ball passes through its maximum height at point A, point B which is at the same height as the cannon, and strikes the ground at point C.

Question

(a) Determine the amount of time that the cannonball is a projectile. (Seconds)
(b) find the components of the ball's velocity at the given points a,b,c, the ball's speed at that point, and the amount of time that elapses before the ball reaches that point.
for the initial time is 0 seconds and the initial speed is 50 m/s, my teacher explained it but I'm still so confused on how to solve it

Answer 1

the height of the ball is
h(t) = 100 + 50 sin53° * t - 4.9t^2
h=0 when t = 10.158

Vy = 50 sin53° - 9.8t
Vx = 50 cos53°
So find t at the points A,B,C
B is the vertex, which is at t = -b/2a = 50 sin53°/9.8
the speed is, of course, √(Vx^2 + Vy^2)

Answer 2

yeah i understand that part, but how am i suppose to get the other points using that if i only know the first velocity which is 50 m/s

Answer 3

oobleck told you:
Vy = 50 sin53° - 9.8t
Vx = 50 cos53°
Vx NEVER CHANGES, (no force in horizontal direction)
You are at the vertex (top) when Vy = 0
the ball hits ground when h = 0, t = 10.158, you can find Vy then and of course you know Vx