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**Hint:**First of all, use the Pythagoras theorem to get the values of $\sin \alpha $ and $\cos \alpha $ from the given value of $\tan \alpha $ . Resolve the components of initial velocity of the ball and calculate the time taken by the ball to get to a height of $15m$ using the formula for position at time $t$ (for y axis). You will get two answers, one for when the ball reaches at a height of $15m$ and other for when the ball continues its projectile motion and again reaches a height of $15m$ before touching the ground. Subtract these two to get the final value.

**Formula Used:**

In a right triangle, ${h^2} = {p^2} + {b^2}$ where $h$ is the hypotenuse of the triangle, $p$ is the perpendicular of the triangle, $b$ is the base of the triangle.

In a projectile motion, position of the object at time $t$ in y-direction is given by $y = ({v_y}\sin \theta )t - \dfrac{1}{2}g{t^2}$ where, ${v_y}$ is the initial velocity of the object in y-direction, $\theta $ is the angle of the projectile motion, $t$ is the time at which position is to be calculated, $g$ is the acceleration due to gravity.

**Complete Step by Step Solution:**

We are given $\tan \alpha = \dfrac{5}{{12}}$ . Since $\tan \theta = \dfrac{p}{b}$ , $p$ is perpendicular, $b$ is base ; we can conclude that perpendicular $ = 5x$ and base $12x$

Using Pythagoras theorem, ${h^2} = {p^2} + {b^2} = {(5x)^2} + {(12x)^2}$ $ \Rightarrow {h^2} = 25{x^2} + 144{x^2} = 169{x^2}$

This gives, $h = 13x$

Therefore, $\sin \alpha = \dfrac{p}{h} = \dfrac{5}{{13}}$ and $\cos \alpha = \dfrac{b}{h} = \dfrac{{12}}{{13}}$

Now, the initial velocity of the ball is given to be $u = 52m{s^{ - 1}}$ at an angle $\alpha $ above the horizontal. Resolve this into its components.

cos component, in horizontal direction is given by ${v_x} = u\cos \alpha = 52 \times \dfrac{{12}}{{13}} = 48m{s^{ - 1}}$

sin component, in vertical direction is given by ${v_y} = u\sin \alpha = 52 \times \dfrac{5}{{13}} = 20m{s^{ - 1}}$

Now we know that in a projectile motion, position of the object at time $t$ in y-direction is given by $y = ({v_y}\sin \theta )t - \dfrac{1}{2}g{t^2}$

We can fix the distance in y-direction in this formula as $15m$ and get the time it took the ball to reach at that height.

(we are considering distance in y-direction and hence only y-direction component of velocity will be used)

On putting respective values, we get $15 = 20 \times t - \dfrac{1}{2} \times 10 \times {t^2}$

Or, \[5{t^2} - 20t + 15 = 0\]

Factorising, \[5{t^2} - 20t + 15 = 0\]

On solving this, we get two values of $t$ i.e. $t = 1\sec $ and $t = 3\sec $

This means that the ball will be at a height of $15cm$ twice in the entire projectile motion.

Once when it is just launched into air, at $t = 1\sec $ and other when it is landing and about to touch land, at $t = 3\sec $

Differences between these times will give us the time when the ball was above $15cm$ . That will be our answer.

Therefore, the final answer will be $3 - 1 = 2\sec $

**Hence, option (B) is correct.**

**Note:**We have taken the values of $p,\,\,b,\,\,h$ as a multiple of $x$ because $\tan ,\,\sin ,\,\cos $ are just a ratio of the values of length of sides of a right triangle. The real values can be anything and hence cannot assign just any value to its side just because it is not relevant to the question. Now $x$ can be 1, or any other number and hence the equations will be true for all values of such triangles.

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