3-2.Motion in Plane
hard

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 $g =10\, ms ^{-2}$ )

A

$20$

B

$25$

C

$29$

D

$200$

(JEE MAIN-2022)

Solution

$v \cos \alpha=u \cos 45^{\circ}$

$v \sin \alpha=u \sin 45^{\circ}-g t$

Solve for $u$ we get

$u=20 \sqrt{2} m /s$

$\Rightarrow H=\frac{u^{2} \sin ^{2} 45^{a}}{20}=20 m$

Standard 11
Physics

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