Question

A body thrown vertically upwards crosses a point ‘p’ on its path after 2 seconds during its upward

journey and again crosses the same point after 6 seconds during downward journey. Find the

height of the point ‘p’ from the ground (as a multiple of acceleration due to gravity)​

Answer 1

Answer:

Time for reaching the maximum height=2 s

Let initial velocity (at the height of 25 m) be u.

Using the equation of motion

‘V=u+gt’

V=0, t=2 s, and. g=-10 m/s^2

u=gt=10×2=20m/s

Applying conservation law of energy

(h=25 m, u=20 m/s; V=?)

(1/2)m u^2+mgh=(1/2) mV^2

u^2+2gh=V^2

V^2=20^2+2×10×25=900

V=30m/s

Answer 2

Answer:

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Explanation:

height = 6g

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