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Q.

A stone is thrown in vertically upward direction with a velocity of $6 m s^{- 1}$ . Also given, the acceleration of the stone during its motion is $10 m s^{- 2}$ in the downward direction; calculate the height attained by it and the time taken to reach there.

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a

1.8 m

and

0.6 s

b

0.5 m

and

1.25 s

c

2.3 m

and

0.32 s

d

11.6 m

and

0.45 s

answer is A.

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Detailed Solution

The height attained or the distance is

1.8 m

and time taken to reach the top is

0.6 s

Given:
The initial velocity is

u = 6 m s^{- 1}

.
The final velocity

v = 0

(the stone stops at the top)
Acceleration

a = - 10 cm s^{- 2}

(Since the stone is thrown vertically upwards, the velocity of the stone goes on decreasing. This is due to the force of gravity of the earth that is acting on it in the downward direction. So, the acceleration is negative.)
To find the distance (or height) and time
Using the third equation of motion formula

v^{2} = u^{2} + 2 as

…(1)
Where

v, u, s,

and

a

are final velocity, initial velocity, displacement, and acceleration, respectively.

{\Rightarrow 0}^{2} = 6^{2} + 2 (- 10) s

\Rightarrow 20 s = 36

\Rightarrow s = \frac{36}{20} = 1.8 m

\Rightarrow s = 1.8 m

Now by using the first equation of motion,

\Rightarrow v = u + at

…(2)

\Rightarrow 0 = 6 + (- 10) t

\Rightarrow t = 0.6 s

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