A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a) - the initial launch velocity, u, and b) - the angle 0, at which the projectile contacts the ground.

(a) : The launch velocity is Vx = 666.66 m/s.

(b) : The angle wich the projectile contacts the ground is α = 0.38°

Explanation:

h = 1m

g = 9.8 m/s²

h = g*t²/2

t = 0.45 s

Vy = g*t

Vy = 4.42 m/s

d=Vx * t

Vx = 666.66 m/s (a)

α = tg⁻¹ (Vy/Vx)

α = 0.38° (b)