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

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Kinematics

A body is projected from the ground at an angle of

4 5^{\circ}

with the horizontal. Its velocity after 2 s is

20 ms^{- 1}

. The maximum height reached by the body during its motion is ______ m. (use

g = 10 ms^{- 2}

)

Answer: 20

Solution

t = 2

s, the velocity is 20 m/s. Since the body is projected at

4 5^{\circ}

, initially

u_{x} = u_{y} = u cos 4 5^{\circ} = u sin 4 5^{\circ}

.

The horizontal component remains constant:

v_{x} = u_{x}

.

The vertical component at

t = 2

v_{y} = u_{y} - g t = u_{y} - 20

.

Given

∣ v ∣ = 20

v_{x}^{2} + v_{y}^{2} = 20

u_{x}^{2} + (u_{x} - 20)^{2} = 20

2 u_{x}^{2} - 40 u_{x} = 0

, so

u_{x} = 20

m/s.

Therefore

u_{y} = 20

m/s.

Maximum height:

H = \frac{u _{y}^{2}}{2 g} = \frac{( 20 ) ^{2}}{2 ( 10 )} = 20

m.

Answer: 20

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer

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