Question 247 of 308

A body is projected from the ground at an angle of $45^{\circ}$ with the horizontal. Its velocity after $2s$ is $20{ms}^{-1}$. The maximum height reached by the body during its motion is _____ $m$. (use $g=10{ms}^{-2}$)

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Answer: 20

After $2s$ the velocity is $20m{s}^{-1}$
$\Rightarrow {v}_{x}^{2}+{v}_{y}^{2}={v}^{2}=400$
$\Rightarrow {u}^{2}{\cos }^{2}45^{\circ}+{\left(u\sin 45^{\circ}-gt\right)}^{2}=\dfrac{{u}^{2}}{2}+{\left(\dfrac{u}{\sqrt{2}}-20\right)}^{2}=400$
$\Rightarrow \dfrac{{u}^{2}}{2}+\dfrac{{u}^{2}}{2}+400-\dfrac{40}{\sqrt{2}}u=400$
$\Rightarrow {u}^{2}=\dfrac{40}{\sqrt{2}}u\Rightarrow u=\dfrac{40}{\sqrt{2}}m{s}^{-1}$
Now the maximum height of the projectile is given by,
$H=\dfrac{{u}^{2}{\sin }^{2}\theta }{2g} \Rightarrow H=\dfrac{1600}{20\times 2\times 2}=20m$