Question

A Diwali rocket, when ignited at the ground level, rises vertically upwards to the level of a window 10 m from the ground. Find the magnitude of velocity of the rocket at the time of its take off. ​

Answer 1

Answer:10√2 m/sec

Explanation: a= -10m/sec² ( because acceleration is in the opposite direction of motion)

v=0(because at the max height velocity is zero)

By third equation of motion,

v²- u² = 2as

-u²=2×-10×10

u²=200

u=√200

u=10√2 m/sec

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Answer 2

here , height = 10m

so, as we know that h= u² /2g

gere g = 10m/s²

10 = u² / 20

u² = 200

u = 10√2 m/s

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