Question

An arrow is shot in air; its time of flight is 5 s and horizontal range is 200 m. The angle that the arrow made with the horizontal will be?
A- \[{{\tan }^{-1}}\dfrac{5}{8}\]
B- \[{{\tan }^{-1}}\dfrac{8}{5}\]
C- \[{{\tan }^{-1}}\dfrac{1}{8}\]
D- \[45{}^\circ \]

Answer 1

Hint: Given is the problem of the projectile, projectile refers to the motion of any body thrown into the air making some angle with the horizontal. In the case of projectile motion, the only force which acts on the body is the force due to gravity.

Complete step by step answer:Given that the total time taken by the arrow to complete the flight is 5 s and the horizontal distance called range covered is 200m
We know the formula for the time of flight is \[t=\dfrac{2u\sin \alpha }{g}\]and the range is given by the formula \[R=\dfrac{{{u}^{2}}\sin 2\alpha }{g}\]
Defining a relationship between the two, we get \[\dfrac{R}{T}=\dfrac{200}{5}=40=u\cos \alpha \]
\[u\cos \alpha =40\]--------(1)
formula for \[R=\dfrac{{{u}^{2}}\sin 2\alpha }{g}\]
\[R=\dfrac{{{u}^{2}}\times 2\sin \alpha \cos \alpha }{g}\]
Using eq (1), \[\dfrac{2u\sin \alpha }{g}=\dfrac{200}{40}=5\]
\[u\sin \alpha =5\]------(2)
Dividing eq (2) by (1) we get,
$
\implies \dfrac{u\sin \alpha }{u\cos \alpha }=\dfrac{5}{40}=\dfrac{1}{8} \\
\implies \tan \alpha =\dfrac{1}{8} \\
$
\[\alpha ={{\tan }^{-1}}\dfrac{1}{8}\]
So, the correct option is (C)

Additional information: Projectile motion is the motion of an object thrown into the air, with some angle with the horizontal and subject to only the acceleration of gravity. There is no acceleration in the horizontal direction and so the horizontal velocity remains constant, the acceleration remains constant throughout the motion of the object.


Note:
This was a good example of using the equations of motions of the projectile while solving the trigonometric equation we have considered the fact the value of the angle is found out in the inverse form. We require a scientific calculator to calculate it.