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

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