Question

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

Answer 1

x (max) = 26760 m

y (max) = 3859 meters

V = 549.5 m/sec

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