A-particle-is-projected-with-velocity-2-gh-so-that-it-just-clears-two-walls-of-equal-heigh-h-in-the-t-1-and-t-2-respectively-The-two-walls-are-at-a-distance-of-2h-from-each-other-If-t

Question Number 154549 by peter frank last updated on 19/Sep/21

A particle is projected with velocity

2 \sqrt{gh} so that it just clears two

walls of equal heigh (h) in the

t_{1} and t_{2} respectively . The two

walls are at a distance of 2 h from

each other . If time passing

between the two walls is 2 \sqrt{\frac{h}{g}}

show that (i) angle projected 60^{°}

(ii) t_{1} + t_{2} = 2 \sqrt{\frac{3 h}{g}}

Commented by Tawa11 last updated on 21/Sep/21

nice sir

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

(i) For horizontal motion between both walls;

x = v_{x} t . But x = 2 h, v_{x} = 2 \sqrt{g h} \cos ϑ, t = 2 \sqrt{\frac{h}{g}}

\Rightarrow 2 h = (2 \sqrt{g h} \cos ϑ) (2 \sqrt{\frac{h}{g}})

\Rightarrow \frac{1}{2} = \cos ϑ \Rightarrow ϑ = 60 °

Commented by peter frank last updated on 20/Sep/21

thanks

Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

(i i) For vertical motion :

y = u t + \frac{1}{2} {a t}^{2}, y = h, u = 2 \sqrt{g h} \sin 60 ° = \sqrt{3 g h}, a = - g

\Rightarrow h = \sqrt{3 g h} t - \frac{1}{2} {g t}^{2} \Rightarrow {g t}^{2} - 2 \sqrt{3 g h} t + h = 0

t = \frac{2 \sqrt{3 g h} \pm \sqrt{12 g h - 4 g h}}{2 g} = \frac{2 \sqrt{3 g h} \pm 2 \sqrt{2 g h}}{2 g}

t_{1} = \sqrt{\frac{3 h}{g}} + \sqrt{\frac{2 h}{g}}, t_{2} = \sqrt{\frac{3 h}{g}} - \sqrt{\frac{2 h}{g}}

t_{1} + t_{2} = 2 \sqrt{\frac{3 h}{g}}

Answered by peter frank last updated on 20/Sep/21

Answered by peter frank last updated on 20/Sep/21

x - direction

x_{1} = v_{o} \cos θ t_{1}

x_{2} = v_{o} \cos θ t_{2}

v_{o} = 2 \sqrt{gh}

x_{2} - x_{1} = 2 h (t_{2} - t_{1}) = 2 \sqrt{\frac{h}{g}}

v_{o} \cos θ (t_{2} - t_{1}) = 2 h

\cos θ = \frac{2 h}{v_{o} (t_{2} - t_{1})} = \frac{1}{2}

θ = \cos^{- 1} \frac{1}{2} = 60^{°}

also y - direction

y = v_{o} \sin θ t_{1} - \frac{1}{2} {gt}_{1}^{2}

y = v_{o} \sin θ t_{2} - \frac{1}{2} {gt}_{2}^{2}

v_{o} \sin θ t_{1} - \frac{1}{2} {gt}_{1}^{2} = v_{o} \sin θ t_{2} - \frac{1}{2} {gt}_{2}^{2}

2 \sqrt{gh} \sin θ (t_{2} - t_{1}) = \frac{1}{2} g (t_{2}^{2} - t_{1}^{2})

2 \sqrt{gh} \sin θ = \frac{1}{2} g (t_{2} + t_{1}) [θ = 60]

t_{2} + t_{1} = \frac{4 \sqrt{gh}}{g} . \frac{\sqrt{3}}{2} =

t_{2} + t_{1} = 2 \sqrt{\frac{3 h}{g}}