Question

A stone is thrown vertically upward from the top of a 59.4m cliff. Ho long is the stone in air. 0.1min 0.6min 0.6sec 4sec​

Answer 1

Concept :-

While going upward body doesn't have initial velocity, it has final velocity only.

Using second equation of Motion,

S = ut + 1/2 gt²

59.4 = 1/2 × 10 × t²

t² = 11.88

t = √11.88

t = 3.44 s

Hence,

The stone was in air = t = 3.44 s

Answer 2

Given : A stone is thrown vertically upward from the top of a 59.4 m cliff .

Need To Find : How long the stone will be in the air ?

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We're provided with the ,Distance Travelled (s) by the body and We'll find Time Taken (t) using Second Equation of motion and that is : $\:\underline {\pmb{\sf s \:=\: ut \:+\:\dfrac{1}{2} + at^2\:}}$

$\twoheadrightarrow \sf s \:=\: ut \:+\:\dfrac{1}{2} + at^2\:\\\\\\\twoheadrightarrow \sf s \:=\: (\: 0\times t\:)\:+\:\dfrac{1}{2} + at^2\:\:\qquad \Bigg\lgroup \:Initial _{\:(Velocity)}\:=\:0\:m/s\:\Bigg\rgroup\:\\\\\\\twoheadrightarrow \sf s \:=\:\:\dfrac{1}{2} + at^2\:\\\\\\\twoheadrightarrow \sf s \:=\:\dfrac{1}{2} + 10 \:\times t^2\:\:\qquad \Bigg\lgroup \:Acc^n\:_{\:(Gravity)}\:=\:10\:m/s^2\:\Bigg\rgroup\\\\\\ \twoheadrightarrow \sf 59.4 \:=\: \dfrac{1}{2} + 10t^2\:\\\\\\\twoheadrightarrow \sf 59.4 \:=\: 5 t^2\\\\\\\twoheadrightarrow \sf 5t^2\:=\:59.4 \\\\\\ \twoheadrightarrow \sf t^2 \:=\: \dfrac{59.4}{5} \\\\\\ \twoheadrightarrow \sf t^2 \:=\: 11.88 \\\\\\\twoheadrightarrow \sf t \:=\: \sqrt{\Big( 11.88\Big)} \\\\\\\twoheadrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:t\:\:=\:3.44\:sec\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf \: Hence,\:Ball\:will\:remain\:in\:air\:for\:\pmb{\sf 3.44\:sec\:}\:.}\:$