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A projectile ...

Question

A projectile is projected at angle $60 °$ with the horizontal. After $2 s$ , velocity vector makes angle $45 °$ with the horizontal. The velocity of projection is $(g = 10 m / s^{2})$

40 (\sqrt{3} + \sqrt{2}) m / s

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\frac{40}{(\sqrt{3} - 1)} m / s

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\frac{20 \sqrt{2}}{\sqrt{3}} m / s

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40 \sqrt{3} m / s

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Solution

The correct option is B $\frac{40}{(\sqrt{3} - 1)} m / s$

$u c o s 60^{o} = u s i n 60^{o} - g \times 2$
$\Rightarrow 20 = u [\frac{\sqrt{3}}{2} - \frac{1}{2}] = \frac{u [\sqrt{3} - 1]}{2}$
$\Rightarrow u = \frac{40}{\sqrt{3} - 1}$

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A projectile is projected at angle 60° with the horizontal. After 2 s, velocity vector makes angle 45° with the horizontal. The velocity of projection is (g= 10 m/s2)

The correct option is B 40(√3−1) m/s u cos 60o=u sin 60o –g×2 ⇒ 20=u[√32−12]=u[√3−1]2 ⇒ u=40√3−1

A projectile is projected at angle $60 °$ with the horizontal. After $2 s$ , velocity vector makes angle $45 °$ with the horizontal. The velocity of projection is $(g = 10 m / s^{2})$

The correct option is B $\frac{40}{(\sqrt{3} - 1)} m / s$

$u c o s 60^{o} = u s i n 60^{o} - g \times 2$
$\Rightarrow 20 = u [\frac{\sqrt{3}}{2} - \frac{1}{2}] = \frac{u [\sqrt{3} - 1]}{2}$
$\Rightarrow u = \frac{40}{\sqrt{3} - 1}$