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# A stone is projected from the ground with a velocity of $14m/s$. One second later it clears a wall $2m$ high. The angle of projection is $\left( {g = 10m{s^{ - 2}}} \right)$A. ${45^ \circ }$B. ${30^ \circ }$C. ${60^ \circ }$D. ${15^ \circ }$

Last updated date: 07th Sep 2024
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Hint First divide the velocity into its two components and then substitute the vertical component in the third equation of motion. Solving this equation we can establish the angle of projection of the stone.
Formula used:
$S = ut - \dfrac{1}{2}g{t^2}$ where $S$ is the vertical distance, $u$ is the velocity, $t$ is the time taken and $g$ is the acceleration due to gravity.

When a stone is projected from the ground with a given velocity, we can divide its velocity into two components- a $\sin$ component which is the vertical component and a $\cos$ component which is the horizontal component with respect to the ground. It is given that the stone clears a wall of height $2m$ so we can use newton’s third equation of motion
$S = ut - \dfrac{1}{2}a{t^2} \\ \Rightarrow 2 = t14\sin \theta - \dfrac{1}{2}g{t^2} \\ \Rightarrow 2 = \left( {14\sin \theta \times 1} \right) - \dfrac{1}{2} \times 10 \times {1^2} \\ \Rightarrow 2 = 14\sin \theta - 5 \\ \Rightarrow 14\sin \theta = 7 \\ \Rightarrow \sin \theta = \dfrac{1}{2} \\ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\ \Rightarrow \theta = {30^ \circ } \\$
Therefore, the angle of projection of the stone with respect to the ground is ${30^ \circ }$ So, the correct option is B.
Note Typically the third equation of motion is described as $S = ut + \dfrac{1}{2}a{t^2}$ but in this particular problem we use the negative sign in this equation as the stone is thrown upwards so the force applied is opposite to the acceleration to the gravity acting on the stone.