Question

7. 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 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

Explanation:

v=u+at

5=0+(10)t

t=1/2

use the formula

s=ut+1/2at²

s=0+1/2x10x(0.5)²

or use the formula

v²-u²=2as

(5)²-0=2(10)(sp

25=20s

s=12.5m friend this is urs answer OK friend bye bye

Answer 2

Given :-

• Initial velocity (u) = 5 m/s

• acceleration (a) = -10 m/s² [because direction of acceleration is opposite to initial velocity].

• Final velocity (v) = 0 m/s

To find :-

• Height attained by the stone and time taken by stone to reach there.

Solution :-

$\underbrace{\bf\purple{{Using\:{3}^{rd}\:equation\:of\:motion}}}$

$\green{\bigstar}{\underline{\boxed{\bf\red{{v}^{2}={u}^{2}+2as}}}}$

$\rm\longrightarrow\:{(0)}^{2}={(5)}^{2}+2\times\:(-10)\times\:s$

$\rm\longrightarrow\:0=25-20s$

$\rm\longrightarrow\:20s=25$

$\rm\longrightarrow\:s=\dfrac{25}{20}$

$\rm\longrightarrow\:s=1.25\:m$

Hence,the hight attained by the stone will be 1.25 m.

Now,

$\underbrace{\bf\green{{Using\:{1}^{st}\:equation\:of\:motion}}}$

$\pink{\bigstar}{\underline{\boxed{\bf\blue{v=u+at}}}}$

$\rm\longrightarrow\:0=5+(-10)\times\:t$

$\rm\longrightarrow\:0=5-10t$

$\rm\longrightarrow\:10t=5$

$\rm\longrightarrow\:t=\dfrac{\cancel{5}}{\cancel{10}}$

$\rm\longrightarrow\:t=\dfrac{1}{2}$

$\rm\longrightarrow\:t=0.5\:sec$

Hence,the time taken by stone to reach there will be 0.5 sec