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Standard XII

Physics

Average Velocity in 2D Motion

A particle is...

Question

A particle is projected horizontally with a speed $10 m s^{- 1}$ at time $t = 0$ from the top of a tower of height $100 m$ . What is the magnitude of tangential acceleration of the particle at time $t = 1 s$ ? $(g = 10 m s^{- 2})$

10 m s^{- 2}

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10 \sqrt{2} m s^{- 2}

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5 \sqrt{2} m s^{- 2}

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5 m s^{- 2}

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Solution

The correct option is C $5 \sqrt{2} m s^{- 2}$
If the angle made by velocity with horizontal is $θ$ , then at $t = 1$
$t a n θ = \frac{V_{y}}{V_{x}} = \frac{g t}{u} = \frac{10 \times 1}{10} = 1 \Rightarrow θ = 45^{0}$

$∴$ Tangential acceleration, $a_{t} = g c o s (90 - θ)$

$∴$ $a_{t} = \frac{g}{\sqrt{2}} = 5 \sqrt{2}$

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A particle is projected horizontally with a speed 10ms−1 at time t=0 from the top of a tower of height 100m. What is the magnitude of tangential acceleration of the particle at time t=1s ? (g=10ms−2)

The correct option is C 5√2ms−2If the angle made by velocity with horizontal is θ, then at t=1tanθ=VyVx=gtu=10×110=1⇒θ=450∴ Tangential acceleration, at=gcos(90−θ)∴ at=g√2=5√2

A particle is projected horizontally with a speed $10 m s^{- 1}$ at time $t = 0$ from the top of a tower of height $100 m$ . What is the magnitude of tangential acceleration of the particle at time $t = 1 s$ ? $(g = 10 m s^{- 2})$