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Question
A stone is thrown vertically upward with a speed of 28 m/s.Find its velocity one second before it reaches the maximum height.
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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)
Total time taken by the stone to reach the maximum height:
\[t = \frac{\left( v - u \right)}{a}\]
\[\Rightarrow t = \frac{\left( 0 - 28 \right)}{- 9 . 8} = 2 . 85 s\]
As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.
t' = 2.85 − 1 = 1.85 s
Again, using the equation of motion, we get:
v' = u + at' = 28 − 9.8 × 1.85
⇒ v' = 28 − 18.13 = 9.87 m/s
Hence, the velocity is 9.87 m/s
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