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26 June, 19:27

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.

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  1. 26 June, 19:39
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    (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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