# Find the time of flight from launch to impact in the following case:A rocket is launched from rest and moves at an acceleration of 25 m/s^2 at an angle of 25 degrees to the horizontal. After 35 s ...

Find the time of flight from launch to impact in the following case:

A rocket is launched from rest and moves at an acceleration of 25 m/s^2 at an angle of 25 degrees to the horizontal. After 35 s of powered flight, the engines shut off and the rocket follows a parabolic path back to earth.

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### 1 Answer

A rocket is launched from rest and moves in a straight line at an angle 25 degrees above the horizontal with an acceleration of 25 m/s^2. After 35 s of powered flight, the engines shut off and the rocket follows a parabolic path back to Earth.

The time of flight of the rocket from launch has to be determined.

When the rocket is powered in addition to the acceleration of 25 m/s^2 there is an acceleration acting vertically downwards of 9.8 m/s^2. When the rocket takes off, the net vertical component of its acceleration is 25*sin 25 - 9.8 = 0.7654 m/s^2. After 35 s of powered flight the height of the rocket is 0 + (1/2)*0.7654*35^2 = 468.84 m.

At a height of 468.84 m the rockets are powered off. The vertical velocity of the rocket is 0.7654*35 = 26.79 m/s. Assuming the gravitational acceleration at this height is also 9.8 m/s^2, the time taken by the rocket to return to the Earth is t where -468.84 = 26.79*t - (1/2)*9.8*t^2

Solving the equation gives the positive value of t = 12.89.

This gives the time of flight from launch till impact as 12.89 + 35 = 47.89 s.