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

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Payal Gupta | Contributor-Level 10

6 months ago

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° * s i n^{2} 45}{2 g}$

$= \frac{40 * 40}{2 * 10} * \frac{1}{2} * \frac{1}{2} = 20 m$

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According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

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