The following graph represents the velocity-time graph of a body projected vertically upward under gravity. Find the maximum height attained by the body. (Take $g = 10 m s^{- 2}$ )

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Solution

At $t = 3 s$ body changes its direction of motion, as its velocity becomes negative from positive.
So maximum height will be area under the velocity time graph.
$h = area$
$\Rightarrow h = \frac{1}{2} \times 30 \times 3$
$\Rightarrow h = 45 m$

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The following graph represents the velocity-time graph of a body projected vertically upward under gravity. Find the maximum height attained by the body. (Take g=10ms−2)

At t=3 s body changes its direction of motion, as its velocity becomes negative from positive.So maximum height will be area under the velocity time graph.h=area⇒h=12×30×3⇒h=45 m

The following graph represents the velocity-time graph of a body projected vertically upward under gravity. Find the maximum height attained by the body. (Take $g = 10 m s^{- 2}$ )

At $t = 3 s$ body changes its direction of motion, as its velocity becomes negative from positive.
So maximum height will be area under the velocity time graph.
$h = area$
$\Rightarrow h = \frac{1}{2} \times 30 \times 3$
$\Rightarrow h = 45 m$