An object is projected with a velocity of $50$ m/s from the ground level at an angle $\theta$ to the vertical. The total time of flight of the projectile is $5$ s. We need to find the value of $\theta$.

1. Problem Description

An object is projected with a velocity of $50$ m/s from the ground level at an angle $\theta$ to the vertical. The total time of flight of the projectile is $5$ s. We need to find the value of $\theta$.

2. Solution Steps

The total time of flight $T$ of a projectile is given by:

$T = \frac{2u_y}{g}$

where $u_y$ is the initial vertical component of velocity and $g$ is the acceleration due to gravity. Here, the projectile is launched at an angle $\theta$ to the vertical. Let the initial velocity be $u = 50$ m/s.

Therefore, the vertical component of initial velocity is $u_y = u \cos \theta$.

Substituting the given values into the equation for total time of flight, we have:

$5 = \frac{2 (50 \cos \theta)}{g}$

Given $g = 10$ m/s²,

$5 = \frac{2 (50 \cos \theta)}{10}$

$5 = \frac{100 \cos \theta}{10}$

$5 = 10 \cos \theta$

$\cos \theta = \frac{5}{10}$

$\cos \theta = \frac{1}{2}$

$\theta = \arccos(\frac{1}{2})$

$\theta = 60^\circ$

3. Final Answer

The value of $\theta$ is $60^\circ$.