Question

A rocket is launched from rest and moves in a straight line at 72.0° above the horizontal with a constant acceleration of 47.8 m/s². After 31.7 s the engines shut off and the rocket follows a parabolic trajectory back to the ground (Assume that the acceleration due to gravity is 9.81 m/s²). What is the time of flight from launch to impact?

What is the maximum altitude reached?

What is the horizontal distance reached?

Answer 1

Displacement = ½ * a * t^2 = ½ * 47.8 * 31.7^2 = 24016 meters.

Final vertical position = 24016.9 * sin 72.0 ° = 22841 m

Final horizontal position = 24016.9 * cos 72.0 ° = 7421 m

vf = vi + (a * t) = 0 + (47.8 * 31.7) = 1515.26 m/s

Final vertical velocity = 1515.2 6* sin 72.0° = 1441 m/s

Final horizontal velocity = 1515.26 * cos 72.0° = 468 m/s

Now, engines shut off and the rocket follows a parabolic path back to earth.

As the rocket moves upward, its vertical velocity decreases 9.8 m/s each second until it reaches its highest vertical position, where its vertical velocity = 0 m/s

Time up = 1441 ÷ 9.8 = 147 seconds

Vertical displacement = (Initial vertical velocity * t) – (½ * 9.8 * t^2)

Vertical displacement = (1441 * 147) – (½ * 9.8 * 147^2)

Vertical displacement = 105943 m

Highest vertical position = 22841 + 105943 =128784 m
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Now the rocket free falls from its highest vertical position to the ground.

Distance = ½ * 9.8 * t^2

128784 = ½ * 9.8 * t^2

t = sqrt(128784 / 4.9)

Time = 162.12 seconds

Time of flight from launch to impact = 31.7 + 147 + 162.12 = 340.82 seconds

During the last 147 + 162.12 = 309.12 seconds, the rocket’s horizontal velocity = 468m/s

Horizontal distance = 468m * 309.12 = 144668 m

Total horizontal distance = 7421 + 144668 = 152089 m