Question

A ball thrown up vertically returns to the thrower after 5 seconds. Find the velocity with which it was thrown up.(acceleration due to gravity= 10m/ sq sec) *

20m/s
22.5m/s
25m/s
29.4m/s

Answer 1

$\large \sf {\underbrace{\underline{Elucidation:-}}}$

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$\rm \red {\underline{Provided\: that:}}$

➻Time taken(t) by the ball to return to thrower= 5 seconds

➻Maximum height (h) of the ball=$\large \sf {\frac{5}{2}=>2.5}$

★Time taken by the ball=2.5 seconds

[∵Maximum height (h) of the ball=2.5]

➻Acceleration due to gravity (a)=$\sf {10m/s^{2}}$

➻As the ball is comming down(de-accelerating) acceleration is denoted with negative sign(retardation).

★Acceleration due to gravity (a)=$\sf {\underline{-10m/s^{2}}}$

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$\rm \blue {\underline{To\: be\: determined:}}$

➻Velocity with which the ball was thrown up,

➻In essence,

➻Initial velocity (u)=?

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$\rm \orange {\underline{Using\: 1^{st}\: equation\: of\: motion:}}$

$\implies {\boxed{v=u+at}}$

➻Since the body deaccelerates and stops,the final Velocity (v) is considered to be "zero".

∴v=0

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➻So supplanting the given values,

$\colon \implies {v=u+at}$

$\colon \implies {0=u+(-10)\times 2.5}$

$\colon \implies {u=10\times 2.5}$

$\colon \implies {u=10\times {\frac{25}{10}}}$

$\colon \implies {u=\cancel{10}\times {\frac{25}{\cancel{10}}}}$

$\colon \implies \green {\fbox{u=25m/s}}$

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$\rm \purple {\underline{Henceforth,}}$

➻The velocity with which it was thrown up,

Intial velocity (u)= 25m/s

★"Option 3 " is the right answer.

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