Question

A ball is dropped from a height $h$. As it bounces off the floor, its speed becomes $80\%$ of what it was just before it hit the floor. The ball then rises to a height equal to

• A. $0.80h$
• B. $0.75h$
• C. $0.64h$
• D. $0.50h$

Answer 1

The correct option is C $0.64h$

According to third equation of motion, $v^2=u^2+2ax$

Here, $u=0m/s$

$a=-g$ and $x=-h$

Negative sign indicates downward direction. Displacement and acceleration both are downwards.

So,

$v=\pm \sqrt{2(-g)(-h)}$

We take minus sign because it is downwards.

$v=-\sqrt{2gh}$

After bouncing. velocity becomes $80\%$ of $v,$ i.e.,
$v^{\prime }=+0.8\sqrt{2gh}$

(positive sign because the direction of ball has reversed after bouncing and is upwards.

Applying third equation of motion again, for $u=v^{\prime }$, $v=0$ and $a=-g$

$v^2=u^2+2\times a\times x$
Thus,
$0=0.64\left(2gh\right)+2(-g)x$
or
$x=0.64h$