Ncert Exercises & Solutions

A ball is dropped from a height of $90~\text m$ on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between $t=0$ to $12~\text s.$

1.		2.
3.		4.

Taking vertical downward motion of ball from a height 90 m, we have
$u = 0, a = 10 m / s^{2}, h = 90 m$
$t = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 \times 90}{10}} = 3 \sqrt{2} s = 4.24 s$
$v = \sqrt{2 a s} = \sqrt{2 \times 10 \times 90} = 30 \sqrt{2} m / s$
$R e b o u n d v e l o c i t y o f b a l l,$
$u' = \frac{9}{10} v = \frac{9}{10} \times 30 \sqrt{2} = 27 \sqrt{2} m / s$
$T i m e t o r e a c h h i g h e s t p o i n t, t' = \frac{u'}{a} = \frac{27 \sqrt{2}}{10} = 3.81 s$
$T o t a l t i m e = t + t' = 4.24 + 3.81 = 8.05 s$
$T h e b a l l w i l l t a k e f u r t h e r 3.81 s t o f a l l b a c k t o f l o o r .$
$V e l o c i t y b e f o r e s t r i k i n g t h e f l o o r = 27 \sqrt{2} m / s$
$V e l o c i t y o f b a l l a f t e r s t r i k i n g t h e f l o o r$
$= \frac{9}{10} \times 27 \sqrt{2} = 24.3 \sqrt{2} m / s$
$T o t a l t i m e e l a p s e d b e f o r e u p w a r d m o t i o n o f b a l l = 8.05 + 3.81 = 11.86 s$
The speed-time graph of the ball is represented in the given figure as:

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions