Question

A ball is thrown from the ground to clear a wall \[3\,{\text{m}}\] high at a distance of \[6\,{\text{m}}\] and falls \[18\,{\text{m}}\] away from the wall. Find the angle of projection of the ball.
A. \[{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{1}{3}} \right)\]
B. \[{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{2}{3}} \right)\]
C. \[{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{3}{2}} \right)\]
D. \[{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{3}{5}} \right)\]

Answer 1

Hint: Use the formula for the equation of trajectory of the projectile. This equation gives the relation between the X and Y coordinates of the projectile at any time, horizontal range of the projectile and angle of projection of the projectile.

Formulae used:

The equation of trajectory of a projectile in terms of the horizontal range of the projectile is given by
\[y = x\tan \theta \left( {1 - \dfrac{x}{R}} \right)\] …… (1)

Here, \[x\] and \[y\] are the horizontal and vertical components of the projectile and \[R\] is range of projectiles.

Complete step by step answer:The ball is thrown from the ground to clear a wall \[3\,{\text{m}}\] high at a distance of \[6\,{\text{m}}\] and falls \[18\,{\text{m}}\] away from the wall.

The horizontal range \[R\] of the ball is the sum of the distance \[6\,{\text{m}}\] of the wall from point of projection and the distance \[18\,{\text{m}}\] at which the ball falls from the wall.
\[R = 6\,{\text{m}} + 18\,{\text{m}}\]
\[ \Rightarrow R = 24\,{\text{m}}\]

Hence, the horizontal range of the ball is \[24\,{\text{m}}\].

The X-coordinate and Y-coordinate of the projectile at the wall are \[6\,{\text{m}}\] and \[3\,{\text{m}}\] respectively.

Substitute \[3\,{\text{m}}\] for \[y\], \[6\,{\text{m}}\] for \[x\] and \[24\,{\text{m}}\] for \[R\] in equation (1).
\[\left( {3\,{\text{m}}} \right) = \left( {6\,{\text{m}}} \right)\tan \theta \left( {1 - \dfrac{{6\,{\text{m}}}}{{24\,{\text{m}}}}} \right)\]
\[ \Rightarrow 3 = 6\tan \theta \left( {1 - \dfrac{1}{4}} \right)\]

Rearrange the above equation for \[\theta \].
\[\theta = {\tan ^{ - 1}}\left[ {\dfrac{3}{{6\left( {1 - \dfrac{1}{4}} \right)}}} \right]\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{1}{2}}}{{\dfrac{3}{4}}}} \right]\]
\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]

Therefore, the angle of projection of the ball is \[{\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right)\].

Hence, the correct option is B.

Note:This question can also be solved by using the kinematic equation for the displacement of the projectile to calculate the horizontal and vertical component of velocity of projectile and then determine an inverse of the ratio of the vertical and horizontal component of the velocity.