Question

A body is thrown vertically upwards from the ground with velocity of 98 m/s. Calculate
(i) The maximum height reached
(ii) Time taken to reach the highest point.
(iii) Velocity at height of 196m from the point of projection.
(iv) Velocity with which it returns to the ground.
(v) Time taken to reach the ground (g =9.8m/s)​

Answer 1

Given:

$\sf{A \ body \ thrown \ vertically \ upwards \ from}$

$\sf{ground \ with \ initial \ velocity (u) \ 98 \ m \ s^{-1}}$

To find:

$\sf{(i) \ The \ maximum \ height \ reached.}$

$\sf{(ii) \ The \ time \ taken \ to \ reach \ the \ highest}$

$\sf{point.}$

$\sf{(iii) \ Velocity \ at \ height \ 196 \ m \ from \ the}$

$\sf{point \ of \ projection.}$

$\sf{(iv) \ Velocity \ with \ which \ it \ returns \ to \ the}$

$\sf{ground.}$

$\sf{(v) \ Time \ taken \ to \ reach \ the \ ground.}$

Calculation:

$\sf{Note:}$

$\sf{For \ upward \ motion,}$

$\sf{(1) \ Final \ velocity (v)=0}$

$\sf{(2) \ Acceleration \ due \ to \ gravity (a)=-9.8 \ m \ s^{-2}}$

$\sf{For \ downward \ motion,}$

$\sf{(1) \ Initial \ velocity(u)=0}$

$\sf{(2) \ Acceleration \ due \ to \ gravity (a)=9.8 \ m \ s^{-2}}$

____________________________________

$\sf{(i)}$

$\sf{According \ to \ the \ third \ equation \ of \ motion.}$

$\boxed{\sf{v^{2}=u^{2}+2as}}$

$\sf{\therefore{0^{2}=98^{2}+2(-9.8)\times \ s}}$

$\sf{\therefore{s=5\times98}}$

$\sf{\therefore{s=490 \ m}}$

$\sf\purple{\tt{\therefore{The \ maximum \ height \ reached \ is \ 490 \ m}}}$

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$\sf{(ii)}$

$\sf{According \ to \ the \ first \ equation \ of \ motion.}$

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

$\sf{0=98+(-9.8)\times \ t}$

$\sf{\therefore{t=\dfrac{98}{9.8}}}$

$\sf{\therefore{t=10 \ s}}$

$\sf\purple{\tt{\therefore{The \ time \ taken \ to \ reach \ the \ highest}}}$

$\sf\purple{\tt{point \ is \ 10 \ second.}}$

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$\sf{(iii)}$

$\sf{Note:}$

$\sf{Here, \ Final \ velocity \ will \ not \ be \ zero}$

$\sf{because \ body \ has \ not \ reached \ the}$

$\sf{maximum \ height.}$

$\sf{According \ to \ the \ third \ equation \ of \ motion}$

$\boxed{\sf{v^{2}=u^{2}+2as}}$

$\sf{\therefore{v^{2}=98^{2}+2(-9.8)(196)}}$

$\sf{\therefore{v^{2}=9604-3841.6}}$

$\sf{\therefore{v^{2}=5762.4}}$

$\sf{\therefore{v=75.9 \ m \ s^{-1} \ (approx)}}$

$\sf\purple{\tt{\therefore{The \ velocity \ at \ the \ height \ of \ 196 \ m}}}$

$\sf\purple{\tt{from \ projection \ is \ 75.9 \ m \ s^{-1}}}$

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$\sf{(iv)}$

$\sf{According \ to \ the \ first \ equation \ of \ motion}$

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

$\sf{\therefore{v=0+(9.8)(10)}}$

$\sf{\therefore{v=98 \ m \ s^{-1}}}$

$\sf\purple{\tt{\therefore{It \ returns \ with \ the \ velocity \ of}}}$

$\sf\purple{\tt{98 \ m \ s^{-1} \ to \ the \ ground.}}$

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$\sf{(v)}$

$\sf{Note:}$

$\sf{Time \ taken \ to \ reach \ ground \ will \ be}$

$\sf{same \ as \ time \ taken \ to \ reach \ maximum}$

$\sf{height.}$

$\sf\purple{\tt{\therefore{It \ takes \ 10 \ second \ to \ reach \ the \ ground.}}}$