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Byju's Answer

Standard XII

Physics

Vector Component

A particle is...

Question

A particle is projected at an angle of $45^{o}$ from 8m before the foot of a wall just touches the top of a wall and falls on the ground on the opposite sides a distance $4 m$ from it. The height of the wall is:

\frac{2}{3} m

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\frac{4}{3} m

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\frac{8}{3} m

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\frac{3}{4} m

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Solution

The correct option is C $\frac{8}{3} m$
Range of projectile $R = 8 + 4 = 12 m$
Angle of projection $θ = 45^{o}$
Using equation of trajectory, $y = x tan θ (1 - \frac{x}{R})$
where $x = 8 m$
Or $y = 8 \times tan 45 (1 - \frac{8}{12})$
$⟹ y = \frac{8}{3} m$

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A particle is projected at an angle of 45o from 8m before the foot of a wall just touches the top of a wall and falls on the ground on the opposite sides a distance 4 m from it. The height of the wall is:

The correct option is C 83mRange of projectile R=8+4=12 mAngle of projection θ=45oUsing equation of trajectory, y=xtanθ(1−xR)where x=8 mOr y=8×tan45(1−812)⟹ y=83 m

A particle is projected at an angle of $45^{o}$ from 8m before the foot of a wall just touches the top of a wall and falls on the ground on the opposite sides a distance $4 m$ from it. The height of the wall is:

The correct option is C $\frac{8}{3} m$
Range of projectile $R = 8 + 4 = 12 m$
Angle of projection $θ = 45^{o}$
Using equation of trajectory, $y = x tan θ (1 - \frac{x}{R})$
where $x = 8 m$
Or $y = 8 \times tan 45 (1 - \frac{8}{12})$
$⟹ y = \frac{8}{3} m$