Question

At what angle a ball should be thrown with a velocity of $24{\text{ m se}}{{\text{c}}^{ - 1}}$just to cross a wall 16m high at a horizontal distance of 32 meters. Given: $g = 10{\text{ m se}}{{\text{c}}^{ - 2}}$

Answer 1

Projectile motion is a form of motion experienced by an object that is projected near the Earth’s surface and moves along a curved path under the action of gravity. The projectile has a single force that acts upon it, which is the force of gravity.
Here, in the question, we need to determine the angle at which the ball should be thrown so as to achieve a height just enough so as to pass the wall, which is 32 meters away from the point of projection. To start with the solution, we need to use the equation of the projectile motion of the body, which is expressed as: $y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$. Substitute all the given values from the question and determine the value of $\theta$ getting the answer.

Complete step by step answer:
The initial velocity of the ball is $u=24m/s$ the horizontal distance $x = 32m$ and the vertical distance $y = 16m$

Let the angle of projection be $\theta$
We know the equation of trajectory for projectile motion is $y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
Hence by substituting the values in the equation we get:

$$16 = 32\tan \theta - \dfrac{{10 \times {{\left( {32} \right)}^2}}}{{2 \times {{\left( {24} \right)}^2}{{\cos }^2}\theta }} \\\ 16 = 32\tan \theta - \dfrac{{10 \times {{\left( {32} \right)}^2}{{\sec }^2}\theta }}{{2 \times {{\left( {24} \right)}^2}}}{\text{ }}\left[ {{{\sec }^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}} \right] \\\ 16 = 32\tan \theta - 5 \times \dfrac{{1024}}{{576}}{\sec ^2}\theta \\\ 16 = 32\tan \theta - 5 \times \dfrac{{16}}{9}{\sec ^2}\theta \\\ 144 = 288\tan \theta - 80\left( {1 + {{\tan }^2}\theta } \right){\text{ }}\left[ {{{\sec }^2}\theta = 1 + {{\tan }^2}\theta } \right] \\\ 144 = 288\tan \theta - 80 - 80{\tan ^2}\theta \\\ 144 + 80 = 288\tan \theta - 80{\tan ^2}\theta \\\ 224 = 288\tan \theta - 80{\tan ^2}\theta \\\ 28 = 36\tan \theta - 10{\tan ^2}\theta \\\$$

It can be written as

$$10{\tan ^2}\theta - 36\tan \theta + 28 = 0 \\\ 5{\tan ^2}\theta - 18\tan \theta + 14 = 0 - - - - (i) \\\$$

Use quadratic equation formula to find$\tan \theta$,

$$\tan \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ = \dfrac{{18 \pm \sqrt {{{18}^2} - 4\left( 5 \right)\left( {14} \right)} }}{{2 \times 5}} \\\ = \dfrac{{18 \pm \sqrt {324 - 280} }}{{10}} \\\ = \dfrac{{18 \pm \sqrt {44} }}{{10}} \\\ = \dfrac{{18 \pm 6.63}}{{10}} \\\ = 2.463;1.137 - - - - (ii) \\\$$

Taking $\arctan$ to both sides of the equation (ii) as:
$\arctan \left( {\tan \theta } \right) = \arctan \left( {2.463} \right);\arctan \left( {1.137} \right) \\\ \theta = {67.90^0};{48.67^0} \\\$
Hence, the angle at which a ball should be thrown with a velocity of $24{\text{ m se}}{{\text{c}}^{ - 1}}$just to cross a wall 16 meters high at a horizontal distance of 32 meters is $\theta = {67.90^0};{48.67^0}$.

Note:
It is to be noted here that the value of acceleration due to gravity should be used as $g = 10{\text{ m se}}{{\text{c}}^{ - 2}}$ and not as $g = 9.8{\text{ m se}}{{\text{c}}^{ - 2}}$, as the value of the constant has already been given in the question. It is interesting here to note that the value of the $\theta$ should be ${45^0}$ for the ball to make a maximum horizontal displacement, but in this question, the vertical, as well as horizontal limits are already mentioned.