Question Number 54767 by pieroo last updated on 10/Feb/19
$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{cliff}\:\mathrm{20m}\:\mathrm{high}.\:\mathrm{After}\:\mathrm{a}\:\mathrm{time}\:\mathrm{of}\:\mathrm{3}\:\mathrm{s} \\ $$$$\mathrm{it}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{edge}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cliff}\:\mathrm{on}\:\mathrm{its}\:\mathrm{way} \\ $$$$\mathrm{down}.\:\mathrm{Calculate} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{projection} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{the}\:\mathrm{speed}\:\mathrm{when}\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left.\mathrm{c}\right)\:\mathrm{the}\:\mathrm{times}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cliff} \\ $$$$\left.\mathrm{d}\right)\:\mathrm{the}\:\mathrm{time}\:\mathrm{it}\:\mathrm{is}\:\mathrm{15m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left(\mathrm{take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{2}} \right). \\ $$
Commented by pieroo last updated on 10/Feb/19
$$\mathrm{NEED}\:\mathrm{SOME}\:\mathrm{HELP} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19
![a)in 3 seconds the stone reach( maximum height and return to the point of projection) so displacement=0 0=ut−(1/2)gt^2 so u=((gt)/2)=((10×3)/2)=15m/sec b)when it reach at the edge of cliff it acquire velocity=0+10×(3/2)=15(v=u+at t=(3/2)sec) require velocity when hit ground v_g ^2 =15^2 +2×10×20=625 v_g =25m/sec c)10=15×t−(1/2)×10t^2 [h=ut−(1/2)gt^2 ] 2=3t−t^2 t^2 −3t+2=0→(t−1)(t−2)=0 t=1sec and 2 sec 1 sec while moving up and 2sec when it reach same place moving down... d)when it is 15m above ground that means 5 meter below the base of cliff... −5=15t−(1/2)×10×t^2 −1=3t−t^2 t^2 −3t+1=0 t=((3+(√(9−4)))/2)=((3+(√5))/2)sec pls chek answer if not correct pls comment...](Q54776.png)
$$\left.{a}\right){in}\:\mathrm{3}\:{seconds}\:{the}\:{stone}\:{reach}\left(\:{maximum}\:{height}\:{and}\:{return}\right. \\ $$$$\left.{to}\:{the}\:{point}\:{of}\:{projection}\right)\:{so}\:{displacement}=\mathrm{0} \\ $$$$\mathrm{0}={ut}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:{so}\:{u}=\frac{{gt}}{\mathrm{2}}=\frac{\mathrm{10}×\mathrm{3}}{\mathrm{2}}=\mathrm{15}{m}/{sec} \\ $$$$\left.{b}\right){when}\:{it}\:{reach}\:{at}\:{the}\:{edge}\:{of}\:{cliff}\:{it}\:{acquire} \\ $$$${velocity}=\mathrm{0}+\mathrm{10}×\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{15}\left({v}={u}+{at}\:\:{t}=\frac{\mathrm{3}}{\mathrm{2}}{sec}\right) \\ $$$${require}\:{velocity}\:{when}\:{hit}\:{ground} \\ $$$${v}_{{g}} ^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{2}×\mathrm{10}×\mathrm{20}=\mathrm{625} \\ $$$${v}_{{g}} =\mathrm{25}{m}/{sec} \\ $$$$\left.{c}\right)\mathrm{10}=\mathrm{15}×{t}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}{t}^{\mathrm{2}} \left[{h}={ut}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \right] \\ $$$$\mathrm{2}=\mathrm{3}{t}−{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{2}=\mathrm{0}\rightarrow\left({t}−\mathrm{1}\right)\left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$${t}=\mathrm{1}{sec}\:\:{and}\:\mathrm{2}\:{sec} \\ $$$$\mathrm{1}\:{sec}\:{while}\:{moving}\:{up}\:{and}\:\mathrm{2}{sec}\:{when}\:{it}\:{reach} \\ $$$${same}\:{place}\:{moving}\:{down}... \\ $$$$\left.{d}\right){when}\:{it}\:{is}\:\mathrm{15}{m}\:{above}\:{ground}\:{that}\:{means}\:\mathrm{5}\:{meter} \\ $$$${below}\:{the}\:{base}\:{of}\:{cliff}... \\ $$$$−\mathrm{5}=\mathrm{15}{t}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×{t}^{\mathrm{2}} \\ $$$$−\mathrm{1}=\mathrm{3}{t}−{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{3}+\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}{sec} \\ $$$${pls}\:{chek}\:{answer}\: \\ $$$${if}\:{not}\:{correct}\:{pls}\:{comment}... \\ $$
Commented by pieroo last updated on 11/Feb/19
$$\mathrm{it}\:\mathrm{is}\:\mathrm{correct}.\:\mathrm{thanks}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19
$${most}\:{welcome}\:...{moral}\:{boost}... \\ $$