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Standard XII

Physics

Instantaneous Velocity

A stone is pr...

Question

A stone is projected vertically up from the ground with velocity $40 m s^{- 1}$ . The interval of time between the two instants at which the stone is at a height of $60$ m above the ground is: $[g = 10 m s^{- 2}]$

4 s

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6 s

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8 s

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12 s

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Solution

The correct option is A 4 s
$60 = u t - \frac{1}{2} g t^{2}$
$60 = 40 t - 5 t^{2}$
$\Rightarrow 5 t^{2} - 40 t + 60 = 0$
$t_{1} + t_{2} = \frac{40}{5} = 8$
$t_{1} t_{2} = \frac{60}{5} = 12$
$t_{1} = 6, t_{2} = 2$
Hence, interval of time $= t_{1} - t_{2} = 6 - 2 = 4 s .$

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A stone is projected vertically up from the ground with velocity 40ms−1 . The interval of time between the two instants at which the stone is at a height of 60 m above the ground is: [g=10ms−2]

The correct option is A 4 s60=ut−12gt260=40t−5t2⇒5t2−40t+60=0t1+t2=405=8t1t2=605=12t1=6,t2=2Hence, interval of time =t1−t2=6−2=4 s.

A stone is projected vertically up from the ground with velocity $40 m s^{- 1}$ . The interval of time between the two instants at which the stone is at a height of $60$ m above the ground is: $[g = 10 m s^{- 2}]$

The correct option is A 4 s
$60 = u t - \frac{1}{2} g t^{2}$
$60 = 40 t - 5 t^{2}$
$\Rightarrow 5 t^{2} - 40 t + 60 = 0$
$t_{1} + t_{2} = \frac{40}{5} = 8$
$t_{1} t_{2} = \frac{60}{5} = 12$
$t_{1} = 6, t_{2} = 2$
Hence, interval of time $= t_{1} - t_{2} = 6 - 2 = 4 s .$