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A Stone is Thrown Vertically Upward with a Speed of 28 M/S. Find the Maximum Height Reached by the Stone. - Physics

Short Note

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:

\[s = \frac{0^2 - {28}^2}{2\left( - 9 . 8 \right)} = 40 \text{ m }\]

 
  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 3 Rest and Motion: Kinematics
Q 25.1 | Page 52
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