Question

A body is launched from the ground and it follows a parabolic path. The total range of the projectile is $40\;m$. When the body has a horizontal displacement of $10\;m$ its height above the ground is $7.5\;m$. Calculate the angle of the launch.

& A{{.30}^{{}^\circ }} \\\ & B{{.45}^{{}^\circ }} \\\ & C{{.60}^{{}^\circ }} \\\ & D{{.75}^{{}^\circ }} \\\ \end{aligned}$$

Answer 1

A body experiences a projectile motion when the object is projected at an inclination from the ground. Then the body follows a curved path, which is parabolic in nature and is called the ballistic trajectory. Here, we need to calculate the angle of inclination of the projectile motion.

Formula used:
$y=xtan\theta-\dfrac{1}{2}g\dfrac{x^{2}}{u^{2}cos^{2}\theta}$

Complete answer:
Let $u$ be the initial velocity of the body which is projected at an angle $\theta$ from the horizontal. Let $H$ and $R$ be the maximum distances along the x and y-axis respectively, during the time of flight $T$. Then the initial velocity $u$ can be resolved into two components along the x and y axis respectively .We know that the distance covered by the body along the x and y-axis respectively is given as, $R$ and $H$ which is the maximum horizontal range and maximum height attained by the object.

Here, given that $R=40m$ and $y=7.5m$ at$x=10m$
Then from the trajectory of projectile, we have $y=xtan\theta-\dfrac{1}{2}g\dfrac{x^{2}}{u^{2}cos^{2}\theta}$
Then during $y=7.5m$ at $x=10m$, we have, $7.5=10tan\theta-\dfrac{1}{2}g\dfrac{(10)^{2}}{u^{2}cos^{2}\theta}$
When, $y=0$, we have $x=40$, which is the range of the projectile.
Then we have, $40tan\theta=\dfrac{1}{2}g\dfrac{(40)^{2}}{u^{2}cos^{2}\theta}$
$\implies u^{2} sin\theta cos\theta=\dfrac{40^{2}\times g}{2\times 40}$
$\implies u^{2}=\dfrac{20g}{sin\theta cos\theta}$
Then form the above two equations, we get, $7.5=10tan\theta-\dfrac{1}{2}g\dfrac{(10)^{2}}{\dfrac{20g}{sin\theta cos\theta }cos^{2}\theta}$
$\implies 7.5=10tan\theta-\dfrac{5}{2}\dfrac{1}{\dfrac{cos\theta}{sin\theta}}$
$\implies 7.5=10tan\theta-\dfrac{5}{2}tan\theta$
$\implies 7.5=7.5 tan\theta$
$\implies tan \theta=1$
$\implies \theta =tan^{-1} 1$
$\implies \theta=45^{\circ}$

So, the correct answer is “Option B”.

Note:
When the object is on the air, the only force acting on the body is the force due to gravitation, which pulls the object back to the surface at the speed of acceleration due to gravity. Note that the range of the projected body is greater than the distance the body will cover, when dropped from the same height. Here, however, we are using the equation of trajectory of the projectile to solve the problem.