Question

a stone is thrown in a vertically upward direction with a velocity of 5 m s^-2 .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 time will it take to reach there?
​

Answer 1

Given :

➳ Initial velocity = 5m/s

➳ Acc. due to gravity = 10m/s²

To Find :

⟶ Max. height attained by ball.

⟶ Time taken by ball to reach there.

SoluTion :

➠ For a body thrown vertically upward, g is taken negative.

➠ Velocity of ball at highest point = zero

✴ Max. height attained by ball :

$\longrightarrow\tt\:v^2-u^2=2gH\\ \\ \longrightarrow\tt\:(0)^2-u^2=2(-g)H\\ \\ \longrightarrow\tt\:-u^2=-2gH\\ \\ \longrightarrow\tt\:H=\dfrac{u^2}{2g}\\ \\ \longrightarrow\tt\:H=\dfrac{(5)^2}{2\times 10}\\ \\ \longrightarrow\boxed{ \bf{H=1.25\:m}}$

✴ Time taken by ball :

$\longrightarrow\tt\:v=u+at\\ \\ \longrightarrow\tt\:0=u+(-g)t\\ \\ \longrightarrow\tt\:-u=-gt\\ \\ \longrightarrow\tt\:t=\dfrac{u}{g}\\ \\ \longrightarrow\tt\:t=\dfrac{5}{10}\\ \\ \longrightarrow\boxed{\bf{t=0.5\:s}}$

Answer 2

$\sf{\underline{\underline{Given:}}}$

$\sf{\longrightarrow Initial\:velocity,\:(u) = 5m{s}^{-1}}$

$\sf{\longrightarrow Final\:Velocity,\:(v) =0}$

$\sf{\underline{\underline{Since:}}}$

$\sf{ \: \: \: \: \: "u"\:is\:upward\:and\:"a"\: is\:downward,\:it\:is}$$\sf{a\:retarded\:motion.}$

$\sf{\underline{\underline{Therefore:}}}$

$\sf{ \: \: \: \: \: a = -10m{s}^{-2}}$

$\sf{\underline{\underline{To\:Find:}}}$

$\sf{ \: \: \: \: \: (i)\:Height\:attended\:by\:stone,\:(s) = \: ...?}$

$\sf{ \: \: \: \: \: (ii)\:Time\:Taken\:to \: attain\:height,\:(h) = \:..?}$

$\sf{\underline{\underline{SolUTiON:}}}$

$\sf\underline{(i)\:Using\:the\:relation,\:{v}^{2} - {u}^{2} = 2as,\:we\: have,}$

$\sf\red{\implies s = {v}^{2} - \dfrac{{u}^{2}}{2a}}$

$\sf\purple{\implies {(0)}^{2} - \dfrac{{5}^{2}}{2} \times (-10) = 1.25\:m}$

$\sf\underline{(ii)\:Using\:the\:relation,\:v = u + at:}$

$\sf\green{\implies 0 = 5 + (-10)t }$

$\sf\orange{\implies t = \dfrac{5}{10} = 0.5\:s}$

$\sf{\underline{\underline{Hence:}}}$

$\sf{ \: \: \: \: \: Height = 1.25\:m}$

$\sf{ \: \: \: \: \: Time\: taken\:to\:attain\:height = 0.5\:sec}$