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Standard XII

Physics

General Gravity Problem

the horizonta...

Question

the horizontal drift of the particle when it strikes the ground,

1.67 km

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2.67 km

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3.67 km

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4.67 km

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Solution

The correct option is B 2.67 km
$\frac{d x}{d t} = a y y = \frac{1}{2} g t^{2} \frac{d x}{d t} = \frac{1}{2} a g t^{2} x = \int_{0}^{T} \frac{1}{2} a g t^{2} d t = \frac{a g T^{3}}{6} T = t i m e o f f l i g h t = \sqrt{\frac{2 H}{g}} = \sqrt{80} s x = \frac{\sqrt{5} \times 10 \times 80^{3 / 2}}{6} = 2.67 k m$

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Similar questions

A particle is released from a certain height H=400m. Due to wind it gathers a horizontal velocity component v_x=ay where a=√5 s^-1and y is the verticle displacement of particle from the point of release, then find

A) the horizontal drift of particle when it hits the ground.

B) the speed with which particle strikes the ground (g=10m/s²)

Q. A particle is released from a certain height of

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. Due to the wind, the particle gathers a horizontal velocity component of

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[Take

g = 10 {m/s}^{2}

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Q. A particle at a height '

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' from the ground is projected with an angle

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with the same speed but with an angle of

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with horizontal. Find the angle it makes with the horizontal when it strikes the ground:

Q. A particle at a height '

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' from the ground is projected with an angle

30^{\circ}

from the horizontal. It strikes the ground making angles

45^{\circ}

with horizontal. It is again projected from the same point with the same speed but with an angle of

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with horizontal. Find the angle it makes with the horizontal when it strikes the ground.

Q. A particle is released from a certain height of

600 m

. Due to the wind, the particle gathers a horizontal velocity component of

a y

where

a

\sqrt{10} s^{- 1}

and

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is the vertical displacement of the particle from the point of release. What is the horizontal drift of the particle when it strikes the ground?
[Take

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]

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the horizontal drift of the particle when it strikes the ground,

The correct option is B 2.67 kmdxdt=ayy=12gt2dxdt=12agt2x=∫T012agt2dt=agT36T=timeofflight=√2Hg=√80sx=√5×10×803/26=2.67km