Question

Find how high will it rise and how much time will it take to return to its point of projection if a body is thrown up with a velocity 1496
m/s

Answer 1

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

Initial speed of body (u) = 1496 m/s

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = - 9.8 m/s²

We know that , the Newton's first equation of motion is given by

$\large \mathbb{ \fbox{V = U + AT }}$

Substitute the known values , we get

0 = 1496 + (-9.8) × T

T = (-1496)/(-9.8)

T = 152.6 sec (approx)

If , time taken for upward direction = time taken for downward direction

Then , it will take 305.2 sec to return to its point of projection

We know that , the Newton's third equation of motion is given by

$\large \mathbb{ \fbox{ {(V)}^{2} - {(U)}^{2} = 2AS }}$

Substitute the known values , we get

(0)² - (1496)² = 2 × (-9.8) × S

S = (-2238016)/(-19.6)

S = 114184.4 m (approx)

Hence , it will reach 114184.4 m high if it is thrown up with a velocity of 1496 m/s

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