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Question
A stone is thrown vertically upward with a speed of 28 m/s. Find the maximum height reached by the stone.
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Solution
Given:
Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s
When the stone reaches the ground, its final velocity (v) is 0.
Also,
a = g = −9.8 m/s2 (Acceleration due to gravity)
Maximum height can be found using the equation of motion.
Thus, we have:
v2 − u2 = 2 as
\[s = \frac{v^2 - u^2}{2a}\]
On putting respective values, we get:
\[s = \frac{0^2 - {28}^2}{2\left( - 9 . 8 \right)} = 40 \text{ m }\]
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