Question 145: Motion in 2D & Vectors

Question

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

Accepted Answer

Given:
The angle of projection $$	heta = 30^{\circ}$$.
The projectile is at the same height at time $$t_1 = 3 \mathrm{~s}$$ and $$t_2 = 5 \mathrm{~s}$$.
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, $$u_y = u \sin 	heta$$.
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, $$u_y$$:
$$u_y = u \sin 	heta = u \sin(30^{\circ}) = \frac{1}{2}u$$
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 $$80 \mathrm{~m/s}$$.

Motion in 2D & Vectors

A projectile fired at $30^{\circ} to the ground is observed to be at same height at time 3 \mathrm{~s} and 5 \mathrm{~s} after projection, during its flight. The speed of projection of the projectile is ___________ \mathrm{m} ~\mathrm{s}^{-1}. (Given g=10 \mathrm{~ms}^{-2}$ )

Choose the correct answer: