Question

A stone is thrown in a vertically upward direction with a velocity of 5 ms¹. If the acceleration of the
stone during its motion is 10 ms² 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

Answer:

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Answer 2

Here We Have

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

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

$\sf \:\:\:\succ\:Velocity \:at \:the\: highest \:point, \:v = 0 m/s²$

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

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

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

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

$\normalsize\mathtt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:by \:the\:ball,\:then}$

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

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

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

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

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

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