At time t = 0, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure below gives the vertical position, y of the apples versus t, time of the falling, until both apples have hit the roadway. The scaling is set by $t_{s}$ = 2.0 s. With approximately what speed was apple 2 thrown down?

15 m/s

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20 m/s

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25 m/s

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10 m/s

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Solution

The correct option is A
15 m/s

$s = u t + \frac{1}{2} a t^{2}$

$s = \frac{1}{2} \times (- 10) (2)^{2} = - 20$

So height of building = 20m

Second time the apple takes only half the time i.e., 1 sec

So

$- 20 = u (1) + \frac{1}{2} (- 10) (1)^{2}$

$- 20 = u - 5$

$u = - 15 m / s$

The apple is thrown with velocity = 15m/s

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The correct option is A 15 m/s s=ut+12at2 s=12×(−10)(2)2=−20 So height of building = 20m Second time the apple takes only half the time i.e., 1 sec So −20=u(1)+12(−10)(1)2 −20=u−5 u=−15m/s The apple is thrown with velocity = 15m/s