Question:

A ball is dropped on to the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact? (Take $g = 10\, m \,s^{-2}$ )

Updated On: Jul 5, 2022
  • $700\, m \,s^{-2}$
  • $1400\, m \,s^{-2}$
  • $2100\, m \,s^{-2}$
  • $2800\, m \,s^{-2}$
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The Correct Option is C

Solution and Explanation

$\upsilon_{1} = \sqrt{2\,gh} = \sqrt{2 ? 10 ? 10} = \sqrt{200}$ $\upsilon _{2} = -\sqrt{2\,gh} = -\sqrt{2 ? 10 ? 10} = -\sqrt{50}$ So $\Delta\upsilon = \upsilon _{1} - \upsilon _{2} = \sqrt{200}+\sqrt{50} = 3\sqrt{50} = 21$ $\therefore\quad$ Acceleration $= \frac{\Delta \upsilon }{\Delta t } = \frac{21}{0.01} = 2100\,m\,s^{-2}$
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Concepts Used:

Motion in a straight line

The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion. 

Types of Linear Motion:

Linear motion is also known as the Rectilinear Motion which are of two types:

  1. Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
  2. Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.