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Question: A stone is projected from the ground with a velocity of \(14m/s\). One second later it clears a wall...

A stone is projected from the ground with a velocity of 14m/s14m/s. One second later it clears a wall 2m2m high. The angle of projection is (g=10ms−2)\left( {g = 10m{s^{ - 2}}} \right)
A. 45∘{45^ \circ }
B. 30∘{30^ \circ }
C. 60∘{60^ \circ }
D. 15∘{15^ \circ }

Explanation

Solution

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−12gt2S = ut - \dfrac{1}{2}g{t^2} where SS is the vertical distance, uu is the velocity, tt is the time taken and gg is the acceleration due to gravity.

Complete step by step answer
We can solve this problem by using Newton's laws of motion. There are two types of systems in classical mechanics- dynamic and kinematic. Dynamics essentially describes the motion of a system considering the forces acting on the body whereas kinematics describes the motion of a system without considering the action of forces acting on the body. In this particular problem, we are dealing with a dynamic system.
When a stone is projected from the ground with a given velocity, we can divide its velocity into two components- a sin⁡\sin component which is the vertical component and a cos⁡\cos component which is the horizontal component with respect to the ground. It is given that the stone clears a wall of height 2m2m so we can use newton’s third equation of motion
S=ut−12at2 ⇒2=t14sin⁡θ−12gt2 ⇒2=(14sin⁡θ×1)−12×10×12 ⇒2=14sin⁡θ−5 ⇒14sin⁡θ=7 ⇒sin⁡θ=12 ⇒θ=sin⁡−1(12) ⇒θ=30∘  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∘{30^ \circ } So, the correct option is B.

Note Typically the third equation of motion is described as S=ut+12at2S = 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.