# You are working as a student intern for the National Aeronautics and Space Administration (NASA) and your supervisor wants you to perform an indirect calculation of the upward velocity of the Space...

You are working as a student intern for the National Aeronautics and Space Administration (NASA) and your supervisor wants you to perform an indirect calculation of the upward velocity of the Space Shuttle relative to the Earth’s surface just 6.1 seconds after it is launched when it has an altitude of 100 meters. In order to obtain data, one of the engineers has wired a streamlined flare to the side of the shuttle that is gently released by remote control after 6.1 seconds. If the flare hits the ground 8.5 seconds after it is released, what is the upward velocity of the flare (and hence of the Shuttle) at the time of its release? (Neglect any effects of air resistance on the flare.) Note: The flare idea is fictional but the data on a typical Shuttle altitude and velocity at 6.1 seconds are straight from NASA!

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### 1 Answer

When the flare is released, it will have two forces acting on it in opposite directions: one upward due to initial upward velocity and the other downward acceleration due to gravity. As a consequence, it will rise a bit and then start falling to the ground.

Initial upward velocity=u

v = 0(since at the topmost point its velocity will become zero)

g=9.81 m/s^2 (since g is downwards)

Applying `v=u-g*t`

we get time to rise up, `t=(u/g) sec`

Height attained in this time, in addition to the initial height of 100 m, can be obtained by applying

`s = u(u/g)-1/2 g(u/g)^2`

`=u^2/(2g)`

So, total height attained by the flare before coming to a standstill and then beginning to fall is:

`= (100+ u^2/(2g))` m,

Let t be the time taken to reach the ground from the topmost point of its flight.

Applying `s=ut+1/2g*t^2,`

`(100+ u^2/(2g)) = 0*t+1/2 g* t^2` (here, g will be positive)

`rArr t^2 = (2/g)(100+ u^2/(2g))`

`=(200/(g)+ u^2/g^2)`

`rArr t = sqrt(200/(g)+ u^2/g^2)`

Total time of flight of the flare after its release:

`sqrt(200/(g)+ u^2/g^2)+u/g`

By condition of the problem: `sqrt(200/(g)+ u^2/g^2)+u/(g)=8.5`

Upon solving, we get `u/(g) =(8.5^2-200/9.81)/17=3.050744`

`rArr u=3.050744*9.81`

`=29.93` m/s

**Therefore, the upward velocity of the flare at the time of its release was 29.93 m/s.**