A hunter tries to hunt a monkey with a small, very poisonous arrow, blown from a pipe with initial speed v. The monkey is hanging on a branch of a tree at height H above the ground. The hunter is at a distance L from the bottom of the tree. The monkey sees the arrow leaving the blowpipe and immediately loses the grip on the tree, falling freely down with zero initial velocity. The minimum initial speed v of the arrow for hunter to succeed while monkey is in air:
A. $\sqrt {\dfrac{{g({H^2} + {L^2})}}{{2H}}} \\ $
B. $\sqrt {\dfrac{{g({H^2} + {L^2})}}{H}} \\ $
C. $\sqrt {\dfrac{{g{H^2}}}{{\sqrt {{H^2} + {L^2}} }}} \\ $
D. $\sqrt {\dfrac{{2g{H^2}}}{{\sqrt {{H^2} + {L^2}} }}} $
Answer
301.2k+ views
Hint:In order to solve this question, we will use the formula of maximum height and range for a projectile motion and form a relation between height and range with initial speed and then we will solve for minimum initial speed of the arrow.
Formula used:
For a projectile motion range R is given by,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
And height H is given by,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
where u is initial velocity and $\theta $ is angle of projection.
Complete step by step solution:
According to the question, let initial velocity of the arrow is u and angle of projection is $\theta $ so distance L is given by using,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
We have
$L = \dfrac{{{u^2}\sin 2\theta }}{g} \to (i)$
And height H is given by using,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
We get,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \to (ii)$
Divide equation (ii) by (i) we get,
$\dfrac{H}{L} = \tan \theta $
So from the right angle triangle we have,

$\sin \theta = \dfrac{H}{{\sqrt {{H^2} + {L^2}} }} \\ $
Now, put this value in equation (ii) we get
$H = \dfrac{{{u^2}{{(\dfrac{H}{{\sqrt {{H^2} + {L^2}} }})}^2}}}{{2g}} \\
\therefore u = \sqrt {\dfrac{{g({H^2} + {L^2})}}{{2H}}} \\ $
Hence, the correct answer is option A.
Note: It should be remembered that, the basic trigonometric ratios are related as $\tan \theta $ is ratio of opposite side to adjacent side whereas sine of angle is the ratio of opposite side to hypotenuse side of the right angle triangle.
Formula used:
For a projectile motion range R is given by,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
And height H is given by,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
where u is initial velocity and $\theta $ is angle of projection.
Complete step by step solution:
According to the question, let initial velocity of the arrow is u and angle of projection is $\theta $ so distance L is given by using,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
We have
$L = \dfrac{{{u^2}\sin 2\theta }}{g} \to (i)$
And height H is given by using,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
We get,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \to (ii)$
Divide equation (ii) by (i) we get,
$\dfrac{H}{L} = \tan \theta $
So from the right angle triangle we have,

$\sin \theta = \dfrac{H}{{\sqrt {{H^2} + {L^2}} }} \\ $
Now, put this value in equation (ii) we get
$H = \dfrac{{{u^2}{{(\dfrac{H}{{\sqrt {{H^2} + {L^2}} }})}^2}}}{{2g}} \\
\therefore u = \sqrt {\dfrac{{g({H^2} + {L^2})}}{{2H}}} \\ $
Hence, the correct answer is option A.
Note: It should be remembered that, the basic trigonometric ratios are related as $\tan \theta $ is ratio of opposite side to adjacent side whereas sine of angle is the ratio of opposite side to hypotenuse side of the right angle triangle.
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