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Question Number 29360 by NECx last updated on 08/Feb/18

Two stones are thrown up   simultaneously from the edge of  a cliff 200m high with initial  speeds of 15ms^(−1)  and 30ms^(−1) .  Verify that the graph shown  below correctly represents the  time variation of the relative  position of the second stone  with respect to the first.Neglect  air resistance and assume that  the stones do not rebound.Take  g=10m/s^2 . Give the equations of  the linear and curved parts of  the plot.

$${Two}\:{stones}\:{are}\:{thrown}\:{up}\: \\ $$$${simultaneously}\:{from}\:{the}\:{edge}\:{of} \\ $$$${a}\:{cliff}\:\mathrm{200}{m}\:{high}\:{with}\:{initial} \\ $$$${speeds}\:{of}\:\mathrm{15}{ms}^{−\mathrm{1}} \:{and}\:\mathrm{30}{ms}^{−\mathrm{1}} . \\ $$$${Verify}\:{that}\:{the}\:{graph}\:{shown} \\ $$$${below}\:{correctly}\:{represents}\:{the} \\ $$$${time}\:{variation}\:{of}\:{the}\:{relative} \\ $$$${position}\:{of}\:{the}\:{second}\:{stone} \\ $$$${with}\:{respect}\:{to}\:{the}\:{first}.{Neglect} \\ $$$${air}\:{resistance}\:{and}\:{assume}\:{that} \\ $$$${the}\:{stones}\:{do}\:{not}\:{rebound}.{Take} \\ $$$${g}=\mathrm{10}{m}/{s}^{\mathrm{2}} .\:{Give}\:{the}\:{equations}\:{of} \\ $$$${the}\:{linear}\:{and}\:{curved}\:{parts}\:{of} \\ $$$${the}\:{plot}. \\ $$

Commented by NECx last updated on 08/Feb/18

Answered by 33 last updated on 08/Feb/18

    v_(rel)  = 30 − 15 = 15 m/s   ...i  ( taking upward direction to be   positive )  a_(rel ) = 10 − 10 = 0 m/s^2   eq   ...i represents straight line    as one of the stones hit (15 m/s)  v_(rel)  > 30 m/s  a_(rwl)  = 0 − 10 = −10m/s^2   ⇒curved graph starts as the 15m/s  stone hits the ground

$$ \\ $$$$ \\ $$$${v}_{{rel}} \:=\:\mathrm{30}\:−\:\mathrm{15}\:=\:\mathrm{15}\:{m}/{s}\:\:\:...{i} \\ $$$$\left(\:{taking}\:{upward}\:{direction}\:{to}\:{be}\:\right. \\ $$$$\left.{positive}\:\right) \\ $$$${a}_{{rel}\:} =\:\mathrm{10}\:−\:\mathrm{10}\:=\:\mathrm{0}\:{m}/{s}^{\mathrm{2}} \\ $$$${eq}\:\:\:...{i}\:{represents}\:{straight}\:{line} \\ $$$$ \\ $$$${as}\:{one}\:{of}\:{the}\:{stones}\:{hit}\:\left(\mathrm{15}\:{m}/{s}\right) \\ $$$${v}_{{rel}} \:>\:\mathrm{30}\:{m}/{s} \\ $$$${a}_{{rwl}} \:=\:\mathrm{0}\:−\:\mathrm{10}\:=\:−\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow{curved}\:{graph}\:{starts}\:{as}\:{the}\:\mathrm{15}{m}/{s} \\ $$$${stone}\:{hits}\:{the}\:{ground} \\ $$$$ \\ $$

Commented by NECx last updated on 08/Feb/18

according to the text book this is  wrong sir.

$${according}\:{to}\:{the}\:{text}\:{book}\:{this}\:{is} \\ $$$${wrong}\:{sir}. \\ $$

Commented by 33 last updated on 08/Feb/18

plz tell me whats the problem  in this one.

$${plz}\:{tell}\:{me}\:{whats}\:{the}\:{problem} \\ $$$${in}\:{this}\:{one}. \\ $$

Answered by Tinkutara last updated on 08/Feb/18

Commented by Tinkutara last updated on 08/Feb/18

First image http://ibb.co/mArn1H

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