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Q.

A body is projected from the ground at an angle of $45^{o}$ with the horizontal. Its velocity after 2s is 20m/s. The maximum height reached by the body during its motion is____m. $(g = 10 m / s^{2})$

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answer is 20.

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Detailed Solution

$\vec{u} = a \hat{i} + b \hat{j}$
For $θ = 45^{o}, a = b$
$\vec{u} = a \hat{i} + a \hat{j}$
$20 = \sqrt{a^{2} + {(a - 10 t)}^{2}} \Rightarrow 20^{2} = a^{2} + {(a - 20)}^{2} \Rightarrow a = 20$
So, $\vec{u} = 20 \hat{i} + 20 \hat{j} \Rightarrow | \vec{u} | = 20 \sqrt{2}$
$H_{M A X} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{{(20 \sqrt{2})}^{2} {(\sin 45^{o})}^{2}}{2 (10)} = \frac{400 \times 2 \times \frac{1}{2}}{20} = 20 m$

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