Question

A ball is dropped from a height of $90m$ 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 $$t = 0{\text{ }}to{\text{ }}12{\text{ }}s$$$(g = 10m{s^{ - 2}})$.

Answer 1

Collision is the sudden, forceful coming together in direct contact of two bodies, it is also called as impact. There are two types of collision.
They are Elastic collision and Inelastic collision.

Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$, where $u$ is initial velocity of the ball, $v$ is Final velocity of the ball, $a$is acceleration, and $t$ is time taken by the ball.

Complete step by step answer:
It is given in the question stated as, Height $s = 90m$
Initial velocity of the ball, $u = 0$
Final velocity of the ball, $v = ?$
Time taken by the ball to reach the ground, $S = ut + \dfrac{1}{2}a{t^2}$ where $u$ is initial velocity of the ball, $v$ is Final velocity of the ball, $a$ Is acceleration, and $t$ is time taken by the ball

On putting the values and we get,
$\Rightarrow 90 = 0 + \dfrac{1}{2} \times 9.8{t^2}$
Taking square root, $t = \sqrt {18.38} = 4.29$ seconds
The final velocity of the ball as it reaches the ground, $v = u + at$
$u = 0$,$a = 9.8$, $t = 4.29s$
$\Rightarrow v = 0 + 9.8 \times 4.29s$
On simplifying we get
$\Rightarrow v = 42.04m{s^{ - 1}}$ $^{}$
The ball loses it’s one tenth of the velocity during collision,
The rebound velocity, ${u_r} = \dfrac{{9v}}{{10}}$
As, $v = 42.04$
$\Rightarrow \dfrac{{9 \times 42.04}}{{10}}$
On simplify we get,
$\Rightarrow 37.84m{s^{ - 1}}$

The time taken by the ball after rebound and reaches the maximum height,
$V = {u_{r\;}} + a{t_1}$
$\Rightarrow 0 = 37.84 + ( - 9.8){t_1}$
$\Rightarrow t = \dfrac{{ - 37.84}}{{ - 9.8}}$
$\Rightarrow 3.86s$
Total time taken by the ball to reach maximum height,
$T = t + {t_1}$
$\Rightarrow 4.29 + 3.86 = 8.15s$

Now the ball travels to ground in the same direction as it time takes to reaches the maximum height, that is $3.86s$
Final velocity of the ball which is equal to the same speed to go up.
After rebound, Velocity of the ball $v'$ = $\dfrac{{9 \times 37.84}}{{10}}$
= $\Rightarrow 34.05m{s^{ - 1}}$
For the second time total time taken by the ball to rebound,
${T_t} = 8.15 + 3.86 = 12.01s$
Total time taken by the ball to rebound for the second time, ${T_t} = 12.01s$.

Note: As per question, we apply the value $a = 9.8m{s^{ - 2}}$ and $t$ as $0$ for initially because the ball is at rest. Here $s$ represents the seconds. Collision is any event in which two or more objects exert forces on each other in a relatively short time.