Question

A particle is projected from a tower of height 2 m
above ground with speed 3 m/s. Neglect any air
resistance to the particle's motion. Maximum
horizontal distance from the foot of the tower where
the particle can reach is (Take g = 10 m/s2)
(1) 1.7 m
(2) 1.9 m
(3) 2.1 m
(4) 2.3 m​

Answer 1

Given:

Height of tower = 2 m

Speed of particle = 3 m/s

To Find:

Maximum horizontal distance from the foot of the tower where the particle can reach

Answer:

Using equation of trajectory:

$\bf y = x \: tan \theta - \dfrac{g {x}^{2} }{2 {u}^{2} } {sec}^{2} \theta$

For P (x, -2):

$\rm \implies - 2 = x \: tan \theta - \dfrac{g {x}^{2} }{2 {u}^{2} } {sec}^{2} \theta \: \: \: \: ...eq_1$

For x max, $\sf \dfrac{dx}{d \theta} = 0$

By differentiating both sides of $\sf eq_1$ we get:

$\rm \implies \dfrac{d}{d \theta} ( - 2) = \dfrac{dx}{d\theta} tan \theta + x \: {sec}^{2} \theta - \dfrac{g}{ {u}^{2} } x \: {sec}^{2} \theta. \dfrac{dx}{d\theta} - \dfrac{g {x}^{2} }{ {u}^{2} } . {sec}^{2} \theta tan \theta \\ \\ \rm \implies 0 = 0 + x \: {sec}^{2} \theta - \dfrac{g {x}^{2} }{ {u}^{2} } . {sec}^{2} \theta tan \theta \\ \\ \rm \implies \dfrac{g { x}^{ \cancel{2}} }{ {u}^{2} } . \cancel{ {sec}^{2} \theta }tan \theta = \cancel{x } \: \cancel{{sec}^{ 2} \theta} \\ \\ \rm \implies \dfrac{gx}{ {u}^{2} } tan \theta = 1 \\ \\ \rm \implies tan \theta = \dfrac{ {u}^{2} }{gx}$

Now putting this value in $\sf eq_1$ we get:

$\rm \implies - 2 = x \: tan \theta - \dfrac{g {x}^{2} }{2 {u}^{2} } (1 + {tan}^{2} \theta ) \\ \\ \rm \implies - 2 = \cancel{x} \times \dfrac{ {u}^{2} }{g \cancel{x}} -\dfrac{g {x}^{2} }{2 {u}^{2} } (1 + \dfrac{ {u}^{4} }{g ^{2} {x}^{2} } ) \\ \\ \rm \implies - 2 = \dfrac{ {u}^{2} }{g} -\dfrac{g {x}^{2} }{2 {u}^{2} } -\dfrac{ \cancel{g} \cancel{{x}^{2} }}{2 {u}^{2} } \times \dfrac{ {u}^{4} }{g ^{ \cancel{2}} \cancel{{x}^{2} } } \\ \\ \rm \implies - 2 = \dfrac{ {u}^{2} }{g} - \dfrac{g {x}^{2} }{2 {u}^{2} } - \dfrac{ {u}^{2} }{2g} \\ \\ \rm \implies \dfrac{g {x}^{2} }{2 {u}^{2} } = 2 + \dfrac{ {u}^{2} }{2g} \\ \\ \rm \implies {x}^{2} = \dfrac{ \cancel{2} {u}^{2} }{g} \bigg( \dfrac{4g + {u}^{2} }{ \cancel{2}g} \bigg) \\ \\ \rm \implies {x}^{2} = \dfrac{ {u}^{2} }{ {g}^{2} } (4g + {u}^{2} ) \\ \\ \rm \implies x = \sqrt{ \dfrac{ {u}^{2} }{ {g}^{2} } (4g + {u}^{2} )} \\ \\ \rm \implies x = \dfrac{u}{g} \sqrt{4g + {u}^{2} } \\ \\ \rm \implies x = \dfrac{3}{10} \sqrt{4 \times 10 + {3}^{2} } \\ \\ \rm \implies x = 0.3\sqrt{40+ 9 } \\ \\ \rm \implies x = 0.3 \times 7 \\ \\ \rm \implies x = 2.1 \: m$

$\therefore$ Maximum horizontal distance from the foot of the tower where the particle can reach = 2.1 m

Correct Option: $\boxed{\mathfrak{(3) \ 2.1 \ m}}$

Answer 2

R = u√(2H/g)

R = 3 √(2×2/10)

R = 3 √0.4

R = 3 × 0.63

R = 1.89 m

R = 1.9 m