A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3 \mathrm{~s}$ and $5 \mathrm{~s}$ after projection, during its flight. The speed of projection of the projectile is ___________ $\mathrm{m} ~\mathrm{s}^{-1}$.

(Given $g=10 \mathrm{~ms}^{-2}$ )

<p>Given:</p> <ol> <li>The angle of projection $\theta = 30^{\circ}$.</li> <li>The projectile is at the same height at time $t_1 = 3 \mathrm{~s}$ and $t_2 = 5 \mathrm{~s}$.</li> <li>The acceleration due to gravity $g = 10 \mathrm{~m/s^2}$.</li> </ol> <p>We need to find the initial speed of projection, $u$.</p> <p>We can use the following equation to find the vertical displacement, $y$, at any time $t$:</p> <p>$y = u_yt - \frac{1}{2}gt^2$</p> <p>Where $u_y$ is the initial vertical component of the velocity, $u_y = u \sin \theta$.</p> <p>Since the projectile is at the same height at $t_1$ and $t_2$, we can write:</p> <p>$u_yt_1 - \frac{1}{2}gt_1^2 = u_yt_2 - \frac{1}{2}gt_2^2$</p> <p>Substitute the values of $t_1$ and $t_2$:</p> <p>$u_y(3) - \frac{1}{2}(10)(3)^2 = u_y(5) - \frac{1}{2}(10)(5)^2$</p> <p>Now, let&#39;s find the initial vertical component of the velocity, $u_y$:</p> <p>$u_y = u \sin \theta = u \sin(30^{\circ}) = \frac{1}{2}u$</p> <p>Substitute $u_y$ in the equation:</p> <p>$\frac{1}{2}u(3) - \frac{1}{2}(10)(3)^2 = \frac{1}{2}u(5) - \frac{1}{2}(10)(5)^2$</p> <p>Now, simplify and solve for $u$:</p> <p>$3u - 90 = 5u - 250$</p> <p>$2u = 160$</p> <p>$u = 80 \mathrm{~m/s}$</p> <p>The initial speed of projection is $80 \mathrm{~m/s}$.</p>