Question

A ball is thrown with a speed of 20 m si at an elevation angle 45°. Find its time of flight and the horizontal range (take g = 10 m s-?​

Answer 1

Answer:

Time = 2root2sec , Range = 40m

Explanation:

$time \: of \: \: flight (t) = \frac{2usin \alpha }{g} \\ given \: it \: u = 20m {s}^{ - 2} \\ \ \: \: \: \: \: \: \: \: \alpha = {45}^{0} \\ t = \frac{2 \times 20 \times sin {45}^{0} }{10} \\ t = 4 \times \frac{1}{ \sqrt{2} } s \\ t = 2 \sqrt{2} s \: \\ range = \frac{ {u}^{2}sin2 \alpha }{g } \\ range = \frac{20 \times 20 \times sin {90}^{0} }{10} \\ range = 40m$

Hope it will help you.

Answer 2

Answer:

$t_f = 2\sqrt{2}, R = 40m$

Explanation:

$u = 20m/s\\g = 10ms^{-2}\\\theta = 45^\circ\\\\t_f = \dfrac{2u\sin{\theta}}{g}\\t_f = \dfrac{2 \times 20 \times \sin{(45)}}{10}\\\\t_f = \dfrac{40 \times \sin{45}}{10}\\\\\sin{45} = \dfrac{1}{\sqrt{2}}\\\\t_f = \dfrac{40 \times 1}{10}\\\\t_f = \dfrac{40}{10\sqrt{2}}\\\\t_f = 2\sqrt{2}s\\\\R = \dfrac{u^2 \sin{2\theta}}{g}\\R = \dfrac{20^2 \sin{(2\times45)}}{10}\\R = \dfrac{400 \times \sin{90}}{10}\\\\sin{90} = 1\\\\ R = \dfrac{400 \times 1}{10}\\\\R = \dfrac{400}{10}\\\\R = 40m$

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