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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.
Solution
Here, $t=3 s , u=?, v=0, h=?, a=g=-9.8 m s ^{-2}$
$u=v-a t$.
$=0-(-9.8) \times 3$
$=29.4 m s ^{-1}$
$h=u t+\frac{1}{2} g t^{2}$
$=29.4 \times 3+\frac{1}{2} \times(-9.8) \times 3^{2}$.
$=88.2+(-44.1)$
$=44.1 m$
Similar Questions
A truck is moving on a straight road with uniform acceleration. The following table gives the speed of the truck at various instants of time.
| Speed $\left(m s^{-1}\right)$ | $5$ | $10$ | $15$ | $20$ | $25$ | $30$ |
| Time $(s)$ | $0$ | $10$ | $20$ | $30$ | $40$ | $50$ |
Draw the speed-time graph by choosing a convenient scale. Determine from it
$(i)$ the acceleration of truck
$(ii)$ the distance travelled by the truck in $50$ seconds.
The following table show os the positon of three persons between $8.00\, am$ to $8.20\, am$.
| Time | Position (in $km$) | ||
| Person $A$ | Person $B$ | Person $C$ | |
| $8.00 \,am$ | $0$ | $0$ | $0$ |
| $8.05 \,am$ | $4$ | $5$ | $10$ |
| $8.10\, am$ | $13$ | $10$ | $19$ |
| $8.15 \,am$ | $20$ | $15$ | $24$ |
| $8.20\, am$ | $25$ | $20$ | $27$ |
$(i)$ Who is moving with constant speed ?
$(ii)$ Who has travelled maximum distance between $8.00\, am$ to $8.05\, am$ ?
$(iii)$ Calculate the average speed of person $'A^{\prime}$ in $k m h^{-1}$
