A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be $v = 7.6 \hat{i} + 6.1 \hat{j}$ in metre per second ( $\hat{i}$ is horizontal, $\hat{j}$ is upward). Give the approximate answers. To what maximum height does the ball rise?

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9 m

8 m

11 m

10 m

answer is D.

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Detailed Solution

$v_{y} = \sqrt{u_{y}^{2} - 2 g h} ∴ u_{y} = \sqrt{v_{y}^{2} + 2 g h} = \sqrt{(6.1)^{2} + 2 \times 10 \times 9.1} = 14.8 m / s H = \frac{u_{y}^{2}}{2 g} = \frac{(14.8)^{2}}{2 \times 10} \approx 11 m$

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