Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.
Answer:
Explanation:
When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .
It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .
b )
The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it . Hence its acceleration is equal to g in vertically downward direction .
c )
It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .
