We throw a ball horizontally at 10m/s.
(a) after how much time does the trajectory form a 30 degree angle?
(b)after how much time after the throw did the velocity double?
(a) Vx remains 10 m/s.
When the trajectory is 30 degrees (and aimed down), Vy / Vx = tan 30 = 0.5774
Vy = 5.774 m/s = (g/2) t^2
Solve for t.
t = 1.085 seconds
(b) You must be throwing the ball from a large height for the velocity to double.
Require that V^2 = Vx^2 + Vy^2 = 4 Vx^2
(so that V = 2 Vx)
Vy^2 = 3 Vx^2 = 300
Vy = 17.32 m/s
t = sqrt(2 Vy/g) = 1.880 s
The ball was thrown from a height of 17.32 m or higher; otherwise it would have hit the ground before the speed could double