A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity $20 m / s$ .The second stone will overtake the first after travelling a distance of (g= $10 m / s^{2}$ ):

13 m

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15 m

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11.25 m

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19.5 m

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Solution

The correct option is C $11.25 m$
After time t, distance covered by first stone is $s = \frac{1}{2} g t^{2}$
After time (t-1), distance travelled by second stone is $s = u (t - 1) + \frac{1}{2} g {(t - 1)}^{2} = 20 t - 20 + \frac{1}{2} g {(t - 1)}^{2}$
These distance must be equal at the time of overtaking,
Thus, $\frac{1}{2} g t^{2} = 20 t - 20 + \frac{1}{2} g {(t - 1)}^{2}$
or, $0 = 20 t - 20 + \frac{1}{2} g (1 - 2 t)$
which gives, $20 t - 20 + 5 - 10 t = 0$ (putting $g = 10$ )
or $t = 1.5$
Put this t in equation for distance to get $s = \frac{1}{2} g t^{2} = \frac{1}{2} 10 {(1.5)}^{2} = 11.25 m$

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A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20m/s.The second stone will overtake the first after travelling a distance of (g= 10m/s2):

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity $20 m / s$ .The second stone will overtake the first after travelling a distance of (g= $10 m / s^{2}$ ):