Question

A ball is dropped on 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 sec, the average acceleration during contact is​

Answer 1

$\mathfrak{\underline{\underline{\large{Solution:-}}}}$

$\sf{\blue{2100m /{sec}^{2} \: upwards}}$

$\mathfrak{\underline{\underline{\large{Explanation:-}}}}$

Velocity at the time of striking the floor,

$\sf{\red{u = \sqrt{2gh} = \sqrt 2 \times 9.8 \times 10 = 14m/s}}$

(- ve since downwards)

$\sf{\red{u = \sqrt{2gh2} = \sqrt2\times 9.8 \times 2.5 = 7m/s}}$

(+ ve since upwards)

$\bf{\orange{Therefore \: change\: in \: velocity}}$

$\sf{\green{v = 7 -(-14) = 21m/s}}$

$\sf{\green{ \frac{ v}{t } = \frac{21}{0.01} = 2100m/{s}^{2} }}$

( upwards since positive)

Answer 2

$\huge\underline\mathfrak\purple{Solution}$