Question

A ball is thrown from a point $A$ with a speed $u$ at an angle $\theta$ with the horizontal. If $g$ is the acceleration due to gravity, the total time taken for it to reach a point $B$ on the same horizontal plane is

• A. $\frac{2u \sin \, \theta}{g}$
• B. $\frac{u \sin^2 \, \theta}{g}$
• C. $\frac{2u^2 \sin \, \theta}{g}$
• D. $\frac{u \sin \, \theta}{g}$

Answer 1

Answer: (A)

$\frac{2u \sin \, \theta}{g}$

Answer 2

Given, initial velocity = $u$,
projection angle = $\theta$
Let time taken to reach at $B$ is $T_f$ here $y = 0$
Using kinematic equation along $y$-axis
$y = u_y T_f + \frac{1}{2} a_y T^2_f$
$0 = u\, \sin\, \theta \, T_f + \frac{1}{2} ( -g)T_f^2$
$\Rightarrow \:\:\:\: T_f = \frac{2 u \, \sin \, \theta}{g}$
So, the correct option is (A) : $\frac{2u \sin \, \theta}{g}$