Question

stone is thrown in a vertically upward direction with a velocity of 5 m s-¹. if the acceleration of the stone during its motion is 10 m s -² in the downward direction, what will be the height attained by the stone and how much will it take to reach there?​

Answer 1

Question -

A stone is thrown in a vertically upward direction with a velocity of 5 $m\: s^{-¹}$. if the acceleration of the stone during its mot-ion is 10 $m\: s^{-2}$ in the downward direction, what will be the height attained by the stone and how much will it take to reach there?

$\:\:$

${\large{\frak{\pmb{\underline{Here \:we\: have, }}}}}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\succ\: Initial\: velocity,\: u = 5 m/s$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\succ\: Acceleration,\: a = -10 m/s²$

$\sf \:\:\:\succ\:Velocity \:at \:the\: highest \:point, \:v = 0 m/s$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\succ\: Height, \:i.e. \:Distance, \:s = ?$

$\sf \:\:\:\:\:\succ\:Time \:(t) \:taken \:to\: reach \:the\: height = ?$

$\:$

${\large{\frak{\pmb{\underline{Now, }}}}}$

$\normalsize\mathtt{(a)\:If\: h\:is\:the\: maximum\: height\: reached}$$\normalsize\mathtt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:by \:the\:ball,\:then}$

$\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\dag\:\underline{\boxed{\sf{\frak{\red{2as\:= v^2-u²}}}}}$

$\:$

$\sf \:\:\:or\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: s\:=\:\bf\dfrac{v²-u²}{2a}$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\bf\dfrac{(0\:m/s)²-(5\:m/s)²}{2×(-10\:m/s²)}$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:s=\bf\dfrac{-25\:m²/s²}{-20\:m/s}$

$\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{=\frak{\green{1.25\:m}}}}}$

$\:$

${\normalsize{\frak{\pmb{\underline{ Thus,\:the\:ball\:will\: reach\:a\: height\:of\:1.25\:m.}}}}}$

$\:\:$

$\normalsize\mathtt{(b)\:If\:t\:is\:the\:time\:of\:ascent,\:then}$

$\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ddagger\:\underline{ \boxed{ \sf t= {\bf\dfrac{v-u}{a} }}}$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\bf\dfrac{0\:m/s-5\:m/s}{-10\:m/s²} = \bf\dfrac{1}{2}\:s$

$\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{=\frak{\orange{O.5\:s}}}}}$

$\:$

Know more

${\bold \star{ \underline{\normalsize{ Notation}}}}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\diamond\:Distance \:\:\:\:\:\:\:\:s\:or\:d$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\diamond\: Initial\:speed\:or\: velocity \:\:\:\:\:\:\:\:u$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\diamond\:Final\:speed\:or\: velocity \:\:\:\:\:\:\:\:v$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\diamond\:Time \:\:\:\:\:\:\:\:t$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\diamond\:Displacement \:\:\:\:\:\:\:\:x\:or\:d$

$\:$

${\bold \star{ \underline{\normalsize{ Important\: formulae}}}}$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:Distance \:\:=\:\:speed×Time$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{ \boxed{ \sf{s=v×t}}}$

$\:\:$

$\sf \:\:\:\:\:\:\:\:Displacement \:\:=\:\: velocity×Time$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{ \boxed{ \sf{x=v×t}}}$

$\:\:$

$\sf Average\: velocity\:\:=\:\: \bf\dfrac{Initial\: velocity+Final\: velocity}{2}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{ \boxed{ \sf{v_{av}\:=\:\bf\dfrac{u+v}{2}}}}$

$\:\:$

$\sf \:\:\:Acceleration\:\:=\:\: \bf\dfrac{Change\:in\: velocity}{Time\:taken}$

$\sf \:\:\:\:=\:\: \bf\dfrac{(Final\: velocity-Initial\:velocity)}{Time\:taken}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{ \boxed{ \sf {a=\:\bf\dfrac{v-u}{t}}}}$

$\:\:$

$\sf \:\:Distance\: travelled\:\:=\:\: Average\:speed×Time$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{ \boxed{ \sf {s=v_{av}×t=\:\bf\dfrac{(u+v)}{2}×t}}}$

Answer 2

$\textbf \purple{ Refer to the attachment }$