F.5 phy projected motion

1. Consider the vertical motion from X to Y,
Use equation: v^2 = u^2 + 2as
with u = -10.sin(30) m/s = -5 m/s, a = -g(=-10 m/s^2), s = -10 m, v =?
hence, v^2 = (-5)^2 + 2.(-10).(-10)
v = -15 m/s

After rebound elastically at Y, the rebound (vertical) speed = 15 m/s
Use equation: s = ut + (1/2)at^2
with s = 0 m, u = 15 m/s, a = -g(=-10), t =?
henc, 0 = 15t + (1/2)(-10)t^2
i.e. 5t^2 = 15t
t = 3 s

Thus, time of flight from Y to Z = 3 s
Horizontal component of velocity of sphere = 10.cos(30) m/s = 8.66 m/s
Therefore, horizontal distance = 8.66 x 3 s = 26 m

2. Horizontal distance between A and B = 500.sin(θ)
Horizontal velocity component of bullet = 625.cos(90-θ) = 625.sin(θ)
Hence, 500.sin(θ) = [625.sin(θ)].t
where t is the time of flight of the bullet.
i.e. t = 500/625 s = 0.8 s

Consider the vertical motion of the bullet.
Vertical velocity component = 625.sin(90-θ) = 625.cos(θ)
Use equation: s = ut + (1/2)at^2
with u = 625.cos(θ), t = 0.8 s, a =-g(=-10 m/s^2), s =?
hence, s = [625.cos(θ)].(0.8) + (1/2).(-10)(0.8^2)
i.e. s = 500.cos(θ) - 3.2

Vertical distance of B above A = 500.cos(θ)
Thus, h = 500.cos(θ) - s = 500.cos(θ) - [500.cos(θ) - 3.2]
i.e. h = 3.2 m