A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
0.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 15.0 m/s2. In Stage 2, the acceleration is 10.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.1) Calculate the maximum altitude.
2) Calculate time required to return to the surface (i.e. the total time of flight).
2) Calculate time required to return to the surface (i.e. the total time of flight).
1 answer:
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Answer:
She must be launched with a speed of 74.2 m/s.
Explanation:
Hi there!
The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:
x = v0 · t · cos θ
vy = v0 · sin θ + g · t
x = horizontal distance traveled at time t.
v0 = initial velocity.
t = time.
θ = launching angle.
vy = vertical component of the velocity vector at time t.
g = acceleration due to gravity (-9.8 m/s²).
To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).
When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:
x = v0 · t · cos θ
Solving for v0:
v0 = x/ (t · cos θ)
Replacing v0 in the second equation:
vy = v0 · sin θ + g · t
0 = x/(t·cos(56°)) · sin(56°) + g · t
0 = 260 m · tan (56°) / t - 9.8 m/s² · t
9.8 m/s² · t = 260 m · tan (56°) / t
t² = 260 m · tan (56°) / 9.8 m/s²
t = 6.27 s
Now, let's calculate v0:
v0 = x/ (t · cos θ)
v0 = 260 m / (6.27 s · cos(56°))
v0 = 74.2 m/s
She must be launched with a speed of 74.2 m/s.
