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Standard XII

Physics

Finding Average Velocity

A particle is...

Question

A particle is projected with speed $u$ at angle $θ$ with horizontal from ground. If it is at same height from ground at time $t_{1}$ and $t_{2}$ , then its average velocity in time interval $t_{1}$ to $t_{2}$ is

Zero

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u sin θ

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u cos θ

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\frac{1}{2} [u cos θ]

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Solution

The correct option is C $u cos θ$
Average velocity $= \frac{t o t a l d i s p l a c e m e n t i n^{'} x^{'}}{t o t a l t i m e}$
$= \frac{u cos θ (t_{2} - t_{1})}{(t_{2} - t_{1})}$
$= u cos θ$

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A particle is projected with speed u at angle θ with horizontal from ground. If it is at same height from ground at time t1 and t2, then its average velocity in time interval t1 to t2 is

The correct option is C ucosθAverage velocity =totaldisplacementin′x′totaltime=ucosθ(t2−t1)(t2−t1)=ucosθ

A particle is projected with speed $u$ at angle $θ$ with horizontal from ground. If it is at same height from ground at time $t_{1}$ and $t_{2}$ , then its average velocity in time interval $t_{1}$ to $t_{2}$ is

The correct option is C $u cos θ$
Average velocity $= \frac{t o t a l d i s p l a c e m e n t i n^{'} x^{'}}{t o t a l t i m e}$
$= \frac{u cos θ (t_{2} - t_{1})}{(t_{2} - t_{1})}$
$= u cos θ$