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A diwali rocket moves vertically with constant acceleration of 5 ms–2. After some time its fuel exhausted and then it falls freely. If maximum height attained by the rocket is 60m. Then find its speeds when the fuel is just exhausted. (g = 10ms–2)

Let h be the vertical distance travelled by rocket before exhausting the fuel.

Speed v of rocket after moved by a vertical distance h is determined from the following equation of motion

" v = u2 + ( 2 a S )  "  ...........................(1)

where u = 0 is initial speed because rocket starts from rest, a = 5 m/s2 is acceleration
and S = h is distance travelled.

v2 = 2 × 5 × h     or    v =   m/s

If total vertical distance travelled by rocket is 60 m , distance travelled vertically in gravitational field is (60-h) m .

When the rocket reaches maximum height , its final speed v = 0 . Hence in equation of motion (1) ,

if we substitute v = 0 and initial speed after exhausting fuel   u  =   m/s , acceleration a = -g = -10 m/s2
and S = (60-h) m , then we get

10 h = 2 × 10 × (60-h)

we get h = 40 m from above expression

hence speed of rocket immediately after exhausting fuel , v =  = 20 m/s
Answered by Thiyagarajan K | 21 Aug, 2021, 01:14: PM

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