Question

A rocket is launched to travel vertically upwards with a constant velocity of 20 m/s. After traveling for 35 s the rocket develops a snag and it's fuel supply is cut-off. The rocket then travels like a free body. What is the total height achieved by it? After what time of its launch will it come back to the earth?​

Answer 1

$\huge{\underline{\underline{\mathfrak{Answer}}}}$

S = ut

This is because it is moving at a constant velocity.

After 35 seconds the distance will be :

S = 20 × 35 = 700m

Now from 700 m above the ground now it travels at a free fall.

The initial velocity = 20 m/s

Final velocity = 0

Acceleration due to gravity = 10

Now :

V² = U² - 2gs

0 = 400 - 20S

20s = 400

S = 400/20

S = 20 m

The total height achieved = 700 + 20

= 720 m

The time it takes to drop back to the earth is :

S = ut + 0.5gt²

u = 0

720 = 0.5 × 10 × t²

720 = 5t²

t² = 720/5

t² = 144

t = √144

= 12 seconds

The total time it takes is :

12 + 35 = 47 seconds

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\purple{Answer:-}$

⏩ Upward distance covered by the rocket in first 35 s

= ut = 20 × 35 = 700 m.

When the fuel supply is cut-off:-

u= +20 m/s, g= -10 m/s², v= 0

Therefore,

v² - u² = 2gs or

0 - (20)² = 2 × (-10) × s

or

s= 400/20 = 20 m.

Total height achieved = 700 + 20 = 720 m. Ans.

Also, t= v-u/g

= 0-20/-10 = 2 s

For downward motion of the rocket :-

u= 0, s= -720 m, g= -10 m/s²

As, s= ut + 1/2gt²

Therefore,

-720 = 0 + 1/2 × (-10) t²

or

t² = 144

or

t= √144 = 12 s

Therefore, total time taken = 35 + 2 + 12 = 49 s. Ans.

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