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A small rocket is fired straight up. At a height of 5.0 km and a velocity of 200.0 m/s, it releases its booster and enters free fall. Calculate the total time of flight between releasing its booster and falling back to the ground.

A. 20.4 s
B. 40.8 s
C. 37.9 s
D. 58.3 s

Answer :

The total time of flight between releasing its boosters and hitting the ground is 40.8s. therefore, B. 40.8s is correct option.


To calculate this, we need to know the initial velocity of the rocket, the height of its release point, and the acceleration due to gravity.

The initial velocity of the rocket is 200.0m/s, the height of its release point is 5.0km, and the acceleration due to gravity is 9.8 m/s^2.

We can calculate the time of flight using the equation:

t = 2v/a.

Plugging in the known values, we get:

t = 2 × 200/9.8

= 40.8s.

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