7. MOTION
hard

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 ?

A

$t=1 \,s\;;\;s=1.25 \, m$

B

$t=0.5 \,s\;;\;s=2.25 \, m$

C

$t=1 \,s\;;\;s=2.5 \, m$

D

$t=0.5 \,s\;;\;s=1.25 \, m$

Solution

Given Initial velocity of stone, $u =5\, m s ^{-1}$

Downward of negative Acceleration, $a =10 \,m s ^{-2}$  

We know that $2 \,as = v ^{2}- u ^{2}$

Therefore, Height attained by the stone, $s =\frac{0^{2}}{5^{2}} \times(-10)  \,m$

$=\frac{-25}{-20}  \,m$

$=1.25 \, m$

Also we know that final velocity, $v = u +at$

or, Time, $t =\frac{v-u}{a}$

Therefore, Time, $t$ taken by stone to attain the height, $s=\frac{0-5}{- \,10  \,s}$

$=0.5  \,s$

Standard 9
Science

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