Two boys conducted experiments on the projectile motion with stopwatch and noted some readings. As one boy throws a stone in air at the same angle with the horizontal, the other boy observes that after $4 \mathrm{~s}$, the stone is moving at an angle $30^{\circ}$ to the horizontal and after…

Question

Two boys conducted experiments on the projectile motion with stopwatch and noted some readings. As one boy throws a stone in air at the same angle with the horizontal, the other boy observes that after $4 \mathrm{~s}$, the stone is moving at an angle $30^{\circ}$ to the horizontal and after another $2 \mathrm{~s}$ it is travelling horizontally. The magnitude of the initial velocity of the stone is (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$.)

Aicharya · Accepted Answer

Given, acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ After $4 \mathrm{~s}$, angle between stone and horizontal plane, $\theta=30^{\circ}$ After $t=4 \mathrm{~s}$, equation of the vertical projectile motion, when $\theta=30^{\circ}$

Total time to reach the stone at horizontal surface, $t=2+4=6 \mathrm{~s}$ After $t=6 \mathrm{~s}$, equation of horizontal projectile motion, $\theta=0^{\circ}$, $ \tan 0^{\circ}=\frac{v \sin \theta-g(6)}{v \times \cos 0} \quad\left[\begin{array}{c} \because \cos 0^{\circ}=1 \\ \tan 0^{\circ}=0 \end{array}\right] $ or $v \sin \theta-g(6)=0$

From Eq. (ii), At $t=4 \mathrm{~s}$, when particles travelling in horizontal direction,

Now, magnitude of initial velocity, $ \begin{aligned} & v=\sqrt{(v \sin \theta)^2+(V \cos \theta)^2} \\ & \text { [From Eqs. (iii) and (iv)] } \\ & \text { or } \\ & v=\sqrt{(60)^2+(20 \sqrt{3})^2} \\ & \text { or } \\ & v=40 \sqrt{3} \mathrm{~m} / \mathrm{s} \\ & \end{aligned} $ So, the magnitude of initial velocity of stone is $v=40 \sqrt{3} \mathrm{~m} / \mathrm{s}$

Two boys conducted experiments on the projectile motion with stopwatch and noted some readings. As one boy…

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