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Standard XII

Physics

Body in Water

A particle is...

Question

A particle is thrown horizontally with $20 m / s$ from the top of a tower of height $20 m (g = 10 m / s^{2})$ . Angle with horizontal at which it will hit the ground is:-

30^{\circ}

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60^{\circ}

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53^{\circ}

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45^{\circ}

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Solution

The correct option is D $45^{\circ}$
$(D)$
$v_{x} = u_{x} = 20$
using: $v^{2} - u^{2} = 2 a s$ to find $v_{y}$
$v_{y} = \sqrt{2 g h}$
$v_{y} = 20$
$t a n θ = \frac{v_{y}}{v_{x}}$
$t a n θ = 20 / 20 = 1$
$θ = 45^{o}$

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A particle is thrown horizontally with 20m/s from the top of a tower of height 20m(g=10m/s2). Angle with horizontal at which it will hit the ground is:-

The correct option is D 45∘(D)vx=ux=20using: v2−u2=2as to find vyvy=√2ghvy=20tanθ=vyvxtanθ=20/20=1θ=45o

A particle is thrown horizontally with $20 m / s$ from the top of a tower of height $20 m (g = 10 m / s^{2})$ . Angle with horizontal at which it will hit the ground is:-

The correct option is D $45^{\circ}$
$(D)$
$v_{x} = u_{x} = 20$
using: $v^{2} - u^{2} = 2 a s$ to find $v_{y}$
$v_{y} = \sqrt{2 g h}$
$v_{y} = 20$
$t a n θ = \frac{v_{y}}{v_{x}}$
$t a n θ = 20 / 20 = 1$
$θ = 45^{o}$