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11 June, 13:37

A projectile is fired from ground level with an initial speed of 550 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to gravity is 9.8 m/sec2

What is the range of the projectile? ___ meters

What is the max height of the projectile? __ meters

what is the speed at which the projectile hits the ground? ___m/sec

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  1. 11 June, 15:09
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    x (max) = 26760 m

    y (max) = 3859 meters

    V = 549.5 m/sec

    Step-by-step explanation:

    Equations to describe the projectile shot movement are:

    a (x) = 0 V (x) = V (₀) * cos α x = V (₀) * cos α * t

    a (y) = - g V (y) = V (₀) * sin α - g*t y = V (₀) * sin α * t - (1/2) * g*t²

    a) What is the range of the projectile. α = 30°

    then sin 30° = 1/2 cos 30° = √3 / 2 and tan 30° = 1/√3

    x maximum occurs when in the equation of trajectory we make y = 0

    Then

    y = x*tan α - g*x / 2*V (₀) ²*cos² α

    x*tan α = g*x / 2 * V (₀) ²*cos² α

    By subtitution

    1/√3 = 9.8 * x (max) / 2 * (550) ²*0.75

    (1/√3) * 453750 / 9.8 = x (max)

    x (max) = 453750 / 16.95 meters

    x (max) = 26760 m

    The maximum height is when V (y) = 0

    We compute t in that condition

    V (y) = 0 = V (₀) * sin α - g*t

    t = V (₀) * sin α / g ⇒ t = 550 * (1/2) / 9.8

    t = 28.06 sec

    Then h (max) = y (max) = V (₀) sin α * t - 1/2 g * t²

    y (max) = 550 * (1/2) * 28.06 - (1/2) * 9.8 * (28.06) ²

    y (max) = 7717 - 3858

    y (max) = 3859 meters

    What is the speed when the projectile hits the ground

    V = V (x) + V (y) and t = 2 * 28.06 t = 56.12 sec

    mod V = √ V (x) ² + V (y) ²

    V (x) = V (₀) cos α = 550 * √3/2

    V (x) = 475.5 m/sec V (x) ² = 226338 m²/sec²

    V (y) = 550*1/2 - 9.8 * 56.12 ⇒ V (y) = 275 - 549.98

    V (y) = - 274.98 V (y) ² =

    V = √ 226338 + 75614 ⇒ V = 549.5 m/sec
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