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/sec2What is the range of the projectile? ___ metersWhat is the max height of the projectile? __ meterswhat 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