JEE Main 2022 · 24 Jun · Shift 2 · Physics Q51

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PhysicsInteger+4 / -1Question 51 of 90

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

Your answer

Answer and worked solution

Official answer: 20

Components. At 45°, $u_x=u_y=\dfrac u{\sqrt2}$ . After 2 s, $v_x=\dfrac u{\sqrt2}$ and $v_y=\dfrac u{\sqrt2}-20$ .

Speed 20. $\dfrac{u^2}2+\left(\dfrac u{\sqrt2}-20\right)^2=400$ gives $u^2=20\sqrt2\,u$ , so $u=20\sqrt2$ m s $^{-1}$ .

Height. $H=\dfrac{u^2\sin^245°}{2g}=\dfrac{800\times\frac12}{20}=20$ m.

Answer: 20

Topics:Projectile motion

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 51