QJEEM23Phe91b: JEE Main 2023 Physics - Class 11 JEEM23Phe91b Solution: A projectile fired at to the ground is observed to be at same…

Given:

1. The angle of projection $\theta = 30^{\circ}$ .
2. The projectile is at the same height at time $t_1 = 3 \mathrm{~s}$ and $t_2 = 5 \mathrm{~s}$ .
3. The acceleration due to gravity $g = 10 \mathrm{~m/s^2}$ .

We need to find the initial speed of projection, $u$ .

We can use the following equation to find the vertical displacement, $y$ , at any time $t$ :

$y = u_yt - \frac{1}{2}gt^2$

Where $u_y$ is the initial vertical component of the velocity, $u_y = u \sin \theta$ .

Since the projectile is at the same height at $t_1$ and $t_2$ , we can write:

$u_yt_1 - \frac{1}{2}gt_1^2 = u_yt_2 - \frac{1}{2}gt_2^2$

Substitute the values of $t_1$ and $t_2$ :

$u_y(3) - \frac{1}{2}(10)(3)^2 = u_y(5) - \frac{1}{2}(10)(5)^2$

Now, let's find the initial vertical component of the velocity, $u_y$ :

$u_y = u \sin \theta = u \sin(30^{\circ}) = \frac{1}{2}u$

Substitute $u_y$ in the equation:

$\frac{1}{2}u(3) - \frac{1}{2}(10)(3)^2 = \frac{1}{2}u(5) - \frac{1}{2}(10)(5)^2$

Now, simplify and solve for $u$ :

$3u - 90 = 5u - 250$

$2u = 160$

$u = 80 \mathrm{~m/s}$

The initial speed of projection is $80 \mathrm{~m/s}$ .