JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 1)
The initial speed of a projectile fired from ground is $$\mathrm{u}$$. At the highest point during its motion, the speed of projectile is $$\frac{\sqrt{3}}{2} u$$. The time of flight of the projectile is :
$$\frac{u}{g}$$
$$\frac{2u}{g}$$
$$\frac{u}{2g}$$
$$\frac{\sqrt3u}{g}$$
Explanation
$u \cos \theta=\frac{\sqrt{3}}{2} u$
$$ \Rightarrow $$ $\cos \theta=\frac{\sqrt{3}}{2}$
$$ \Rightarrow $$ $\theta=30^{\circ}$
Time of flight $=\frac{2 u \sin \theta}{g}=\left(\frac{u}{g}\right)$
$$ \Rightarrow $$ $\cos \theta=\frac{\sqrt{3}}{2}$
$$ \Rightarrow $$ $\theta=30^{\circ}$
Time of flight $=\frac{2 u \sin \theta}{g}=\left(\frac{u}{g}\right)$
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