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A stone is thrown vertically upward with an initial velocity of $40 m / s$ . Taking $g = 10 m s^{- 2}$ find the maximum height reached by the stone. What is the total distance covered by the stone?

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80 m, 160 m

80 m, 165 m

40 m, 160 m

0 m, 20 m

answer is A.

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Detailed Solution

A stone is thrown vertically upward with an initial velocity of

40 m / s

. Taking

g = 10 m s^{- 2}

. The maximum height reached by the stone is

80 m

and the total distance covered by the stone is

160 m

.
Given data in the problem,
Initial velocity,

u = 40 m / s

Final velocity,

v = 0

As stone is moving upward. So, the value of gravity,

g = - 10 m s^{- 2}

By the third equation of motion,

v^{2} - u^{2} = 2 gh

Where,

u = initial velocity

v = final velocity

ag = acceleration due to gravity

h = height

By putting all values in the above equation,

\Rightarrow 0 - 40^{2} = 2 \times - 10 \times h

\Rightarrow h = \frac{160}{20}

\Rightarrow h = 80 m

Distance covered by the stone,

D = upward + downward

\Rightarrow D = h + h

\Rightarrow D = 2 h

\Rightarrow D = 2 \times 80

\Rightarrow D = 160 m

The maximum height reached by the stone is

80 m

and the total distance covered by the stone is

160 m

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