Question

An object is projected at an angle $\theta$ with speed $v$. The time in which it lands back on is

• A. $t=\frac{v\cos \left(\theta \right)}{g}$
• B. $t=\frac{2v\sin \left(\theta \right)}{g}$
• C. $t=\frac{g}{v\sin \left(\theta \right)}$
• D. $t=\frac{g}{v\cos \left(\theta \right)}$

Answer 1

The correct option is A

$t=\frac{v\cos \left(\theta \right)}{g}$

Explanation for correct option

Step 1: Given data

1. Projectile with an inclination $\theta$
2. Projected with a velocity $v$

Step 2: Formula used

$v=u+at$ where $v$ is the final velocity $a$ is the acceleration $t$

is the time and $u$ is the initial velocity

Step 3: Find the time of flight

The object is projected along the normal of the inclined surface.

So the angle of projection is $\theta +90°$

Here the initial velocity is $v\sin \left(90°+\theta \right)$

and $a=-g$( acceleration due to gravity) and $t$ is the time of attaining maximum height say $h$

As we know velocity at max height,$v=0$

$$\begin{gathered} \therefore v=u+at \\ \Rightarrow 0=v\sin \left(90°+\theta \right)-gt \\ \Rightarrow t=\frac{v\cos \left(\theta \right)}{g} \end{gathered}$$

The projected object will take the same time to return to the ground.

Therefore, for the time of flight $T$

$$\begin{gathered} \Rightarrow T=2t \\ =\frac{2v\cos \left(\theta \right)}{g} \end{gathered}$$

Hence option B is correct.