Question Number 6164 by sanusihammed last updated on 16/Jun/16
$${A}\:{stone}\:{is}\:{thrown}\:{vertically}\:{upwards}\:{with}\:{velocity}\:\mathrm{20}{m}/{s}. \\ $$$${at}\:{the}\:{same}\:{time}\:,\:{and}\:\mathrm{20}{m}\:{vertically}\:{above}\:.\:{a}\:{second}\:{stone}\:{is} \\ $$$${allowed}\:{to}\:{fall}.\:{After}\:{whst}\:{time}\:{and}\:{at}\:{what}\:{height}\:{do}\:{they}\: \\ $$$${collide}\:?. \\ $$$${take}\:\:{g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 17/Jun/16
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{stone}\:\mathrm{which}\:\mathrm{is}\:\mathrm{falling}\:\mathrm{travels} \\ $$$$\mathrm{distance}\:{x}.\:\mathrm{Then}\:\mathrm{stone}\:\mathrm{thrown}\:\mathrm{upwards} \\ $$$$\mathrm{will}\:\mathrm{travel}\:\mathrm{distance}\:\mathrm{20}−{x}. \\ $$$$\mathrm{20}−{x}={u}\centerdot{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\mathrm{20}{t}−\mathrm{5}{t}^{\mathrm{2}} \:\:\left(\mathrm{1}\right) \\ $$$${x}=\mathrm{0}\centerdot{t}+\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\mathrm{5}{t}^{\mathrm{2}} \:\:\:\left(\mathrm{2}\right) \\ $$$${substituting}\:{value}\:{of}\:{x}\:{from}\:\mathrm{2}\:{to}\:\mathrm{1}. \\ $$$$\mathrm{20}−\mathrm{5}{t}^{\mathrm{2}} =\mathrm{20}{t}−\mathrm{5}{t}^{\mathrm{2}} \\ $$$${t}=\mathrm{1}\:\mathrm{s} \\ $$$${x}=\mathrm{5}\:\mathrm{m}\:\left({from}\:{the}\:{top}\right) \\ $$$$ \\ $$