The figure shows velocity-time graph of a stone thrown vertically upwards till it comes back to the ground. Calculate the maximum height reached by the stone.

40 m

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4 m

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80 m

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160 m

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Solution

The correct option is C $80 m$
Maximum height $=$ Area of AOB $= \frac{1}{2} \times O B \times O A = \frac{1}{2} \times 4 \times 40 = 80 m$

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Q. A stone is thrown vertically upward with an initial velocity of

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. Taking

g = 10 m / s^{2}

, find the maximum height reached by the stone. What's the net displacement and the total distance covered by the stone?

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The figure shows velocity-time graph of a stone thrown vertically upwards till it comes back to the ground. Calculate the maximum height reached by the stone.

The correct option is C 80mMaximum height = Area of AOB =12×OB×OA=12×4×40=80m

The correct option is C $80 m$
Maximum height $=$ Area of AOB $= \frac{1}{2} \times O B \times O A = \frac{1}{2} \times 4 \times 40 = 80 m$