Aprojectile 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)

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