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7. MOTION
hard
A piece of stone is thrown vertically upwards. It reaches its maximum height in $3$ second. If the acceleration of the stone be $9.8\, m s ^{-2}$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards. Find the maximum height attained by it.
Option A
Option B
Option C
Option D
Solution
Here, $t=3 s , u=?, v=0, h=?, a=g=-9.8 m s ^{-2}$
$u=v-a t$.
$=0-(-9.8) \times 3$
$=29.4 m s ^{-1}$
$h=u t+\frac{1}{2} g t^{2}$
$=29.4 \times 3+\frac{1}{2} \times(-9.8) \times 3^{2}$.
$=88.2+(-44.1)$
$=44.1 m$
Standard 9
Science
