Question

A rubber ball is bounced on the floor of a room which has its ceiling at a height of 3.2 m from the floor. The ball hits the floor with a speed of 10 m/s, and rebounds vertically up. If all collisions simply reverse the velocity of the ball, without changing its speed, then how long does it take the ball for a round trip, from the moment it bounces from the floor to the moment it returns back to it?
(Acceleration due to gravity is 10 m/s2.)
1)4 s
2)2 s
3)0.8 s
4)1.2 s

Answer 1

Here the time taken by the ball to go upward from the floor is the same as that taken to go downward after hitting the ceiling.

Let each time be t, so the total time to be found is 2t.

Consider when the ball goes upward from the floor.

Given, the ball hits the floor with a velocity $\sf{10\ m\,s^{-1}.}$ Since all collisions reverse the direction of its velocity but not magnitude,

• Velocity of the ball when it goes up from floor, $\sf{u=10\ m\,s^{-1}.}$

• When the ball hits the ceiling, its displacement is, $\sf{s=3.2\ m}$

• The ball experiences acceleration due to gravity, but since upward motion is considered as positive, $\sf{a=-g=-10\ m\,s^{-2.}}$

Then by second equation of motion,

$\longrightarrow\sf{s=ut+\dfrac{1}{2}\,at^2}$

$\longrightarrow\sf{3.2=10t-\dfrac{1}{2}\cdot10t^2}$

$\longrightarrow\sf{3.2=10t-5t^2}$

$\longrightarrow\sf{5t^2-10t+3.2=0}$

$\Longrightarrow\sf{t=\dfrac{10\pm\sqrt{100-4\times5\times3.2}}{10}}$

$\longrightarrow\sf{t=0.4\ s\quad\quad OR\quad\quad t=1.6\ s}$

The time $\sf{t=1.6\ s}$ is taken, when the ball goes upward, reaches its maximum height, and then attains a downward motion towards the ground (or floor) during which there's also a displacement of $\sf{3.2\ m}$ for the ball from the floor. But this is not possible and is not considered unless there's no ceiling anymore!

Hence the actual time is,

$\longrightarrow\sf{t=0.4\ s}$

And so the total time is,

$\longrightarrow\sf{2t}=\bf{0.8\ s}$