The velocity-time graph of a stone thrown vertically upward with an initial velocity of 30 ms^-1 is shown in the Fig. The velocity in the upward direction is taken as positive and that in the downward direction as negative. What is the maximum height to which the stone rises?

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45 m

30 m

90 m

60 m

answer is B.

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Detailed Solution

Given u= 30 ms^-1.

From Fig, the acceleration a = slop of line AB = $\frac{30 m s^{- 1}}{- 3 s} = - 10 m s^{- 2} .$

The maximum height is reached when final velocity v = 0. Using the values of u, v and a in relation v² - u² = 2as, we have:

$0 - 30 \times 30 = 2 \times (- 10) \times s \Rightarrow s = \frac{30 \times 30}{20} = 45 m$

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