Answer:
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
