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Assertion: In javelin throw, the athlete throws the projectile at an angle more than $45^\circ$.
Reason: The maximum range does not depend upon angle of projection.
(A) Both assertion and reason are the correct answer, reason is the correct explanation for assertion.
(B) Both assertion and reason are correct. The answer reason is the incorrect explanation for the assertion.
(C) Assertion is correct and reason is incorrect.
(D) Both assertion and reason are incorrect.

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Last updated date: 18th Jun 2024
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Answer
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Hint A javelin when thrown follows a path of projectile motion. So, we can use the formula of range in a projectile motion. A javelin thrower needs the maximum range, so we will find the condition to obtain maximum range and whether it depends on the angle of projection or not.

Formula used
Range in a projectile motion, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Here,
Range is the horizontal distance covered by the object in a projectile motion.
Range is represented by $R$
Velocity of the object is represented by $u$
Gravity is represented by $g$
The angle at which the object is thrown with respect to the horizontal is represented by $\theta $

Complete step by step answer:
The path taken by a javelin is the same as that of a projectile motion. So the formula of range in projectile motion is applicable.
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
\[\sin \theta \] has a range of values from $0$ to $1$
For the range to be maximum \[\sin 2\theta \] should be $1$ since \[\sin 2\theta \] is directly proportional to the magnitude of range from the formula of range.
\[\sin 2\theta = 1\]
$ 2\theta = {\sin ^{ - 1}}1 $
 $ \Rightarrow 2\theta = 90^\circ $
 $ \Rightarrow \theta = 45^\circ $
Hence the angle should be equal to $\theta = 45^\circ $
From this, we can say that for maximum range the angle at which the javelin is thrown should be exactly $\theta = 45^\circ $
And it clearly depends on the angle at which it is thrown with respect to the horizontal.

Hence, the correct answer is Option(D) Both assertive and reason are incorrect.

Note We take gravity in the formula for finding range because the only acceleration acting on the object vertically is gravity. In case the question mentions that the acceleration with which the object is pulled down is different then we take that value in place of gravity.