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Question: A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30 m/s at an angle of 3...

A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30 m/s at an angle of 37 above the horizontal. The angle formed by the velocity of the stone with the horizontal when it hits the ground is (take g = 10 m/s2m/{{s}^{2}} )?
A- tan153{{\tan }^{-1}}\dfrac{5}{3}
B- tan134{{\tan }^{-1}}\dfrac{3}{4}
C- tan132{{\tan }^{-1}}\dfrac{3}{2}
D- tan143{{\tan }^{-1}}\dfrac{4}{3}

Explanation

Solution

In this problem, the body is thrown from the top of a tower making some angle with the horizontal. Also, no external force is acting on the system and the only force which is driving the motion of the ball towards the earth is the force of attraction due to gravity.

Complete step by step answer:
Considering the vertical motion first:
Acceleration is constant and is directed downwards.

& s=ut+\dfrac{a{{t}^{2}}}{2} \\\ & s=ut-\dfrac{g{{t}^{2}}}{2} \\\ \end{aligned}$$ Vertical component of initial velocity is $$u\sin {{37}^{0}}$$. Substituting the values, $$\begin{aligned} & 63.8=30\times \sin {{37}^{0}}-5{{t}^{2}} \\\ & 5{{t}^{2}}-18t+63.8=0 \\\ \end{aligned}$$ This is a quadratic equation, upon factorization its root we get t= 5.8 s X component of initial velocity is $$u\cos {{37}^{0}}=30\times \cos 37=24m/s$$. There is no acceleration in horizontal direction. So, horizontal velocity remains constant throughout 24 m/s but there is acceleration in downward direction, so finding out velocity after time t is, $ {{v}_{y}}={{u}_{y}}+at \\\ \implies {{v}_{y}}={{u}_{y}}-gt \\\ \implies {{v}_{y}}=18-10\times 5.8 \\\ \implies {{v}_{y}}=-40m/s \\\ $ So final velocity after 5.8s in vector form can be written as, $$\overrightarrow{v}=24\widehat{i}-40\widehat{j}$$ Let, Angle formed by the velocity of the stone with the horizontal be $$\varsigma $$ $ \varsigma ={{\tan }^{-1}}\dfrac{{{v}_{y}}}{{{v}_{x}}} \\\ \implies \varsigma ={{\tan }^{-1}}\dfrac{40}{24} \\\ \implies \varsigma ={{\tan }^{-1}}\dfrac{5}{3} \\\ $ **So, the correct answer is “Option A”.** **Additional Information:** In projectile problems when the body is projected whether, from the ground or a certain height, there is no acceleration in the horizontal direction. There is constant acceleration in the vertical direction which is always directed downwards and this is constant. This is the acceleration due to gravity. **Note:** The answer comes out in inverse form, we need a scientific calculator to find out the exact value the ball makes with the horizontal. The motion of the body has been splitted into its components and that is possible only because of the fact that acceleration was constant.