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# A golfer standing on level ground hits a ball with velocity of $u = 52m{s^{ - 1}}$ at an angle $\alpha$ above the horizontal. If $\tan \alpha = \dfrac{5}{{12}}$ , then the time for which the ball is at least $15m$ above the ground will be: (take $g = 10m{s^{ - 2}}$ )A) $1\sec$B) $2\sec$C) $3\sec$D) $4\sec$

Last updated date: 17th Jun 2024
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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.