2.7 Applications of Quadratic Equation in Real World

2.7 Applications of Quadratic Equation in Real World

Example 19

A ball is thrown upward with an initial velocity of $20 \text{ ms}^{-1}$. Calculate the maximum height it reaches above ground level. Calculate time of return to the ground.

Solution Maximum height

At its highest point, the ball's velocity is zero. So, we have the following quantities:

Initial velocity $= v_i = 20 \text{ ms}^{-1}$
Final velocity $= v_f = 0 \text{ ms}^{-1}$
Gravitational acceleration $= g = -10 \text{ m s}^{-2}$
Distance (displacement) $= s = ?$

The relevant equation of motion is $v_f^2 = v_i^2 + 2gs$

Substituting values of $v_i, v_f$ and $g$, we get

$0^2 = (20)^2 + 2(-10)s$

$0 = 400 - 20s$

so, $s = \dfrac{400}{20} = 20m$

Maximum Time of Return: When the ball returns to the point from which it was thrown, its displacement ($s$) is zero. We have the following information:

Displacement $= s = 0$
Initial velocity $= v_i = 20 \text{ ms}^{-1}$
Gravitational acceleration $= g = -10 \text{ ms}^{-2}$
Time $= t = ?$

The relevant equation of motion:

$s = v_i t + \dfrac{1}{2}gt^2$

Putting the values of $s, v_i$ and $g$

lightbulb

Note
The equations of motion describe the relationship between displacement, velocity, time and acceleration for objects moving in a straight line with uniform acceleration.
First equation : $v_f = v_i + at$
Second equation : $s = v_i t + \dfrac{1}{2}at^2$
Third equation : $v_f^2 = v_i^2 + 2as$

$0 = 20t + \dfrac{1}{2}(-10)t^2$

$0 = 20t - 5t^2$

$0 = t[20 - 5t]$

Now $t = 0 \text{ sec}$ or $20 - 5t = 0 \Rightarrow t = \dfrac{20}{5} = 4 \text{ sec}$

(i) $t = 0 \text{ sec.}$ means, the ball had zero displacement at the time it was thrown.
(ii) $t = 4 \text{ sec.}$ means, the ball returned to the ground after $4$ sec, it is the return time or the required time.

Example 20

A company models its profit $P$(rupees in thousands) by the equation $P(x) = -2x^2 + 80x - 500$, where $x$ is the price per item (in rupees). Find the price that gives the maximum profit and calculate the maximum profit.

Solution $P(x) = -2x^2 + 80x - 500$

Since $-2 < 0$, the parabola opens downward and vertex gives the maximum point.

$x = \dfrac{-b}{2a} = \dfrac{-80}{2(-2)} = \dfrac{-80}{-4} = 20$

Hence, the price per item that gives maximum profit is Rs. $20$.

$P(20) = -2(20)^2 + 80(20) - 500$

$= -2(400) + 1600 - 500$

$= 300$ (in thousands)

Hence, maximum profit is Rs. $300,000$.

lightbulb

Note
We can also find vertex by using $x = \dfrac{-b}{2a}, y = f\left(\dfrac{-b}{2a}\right)$

Example 21

A ball is thrown upwards from the ground. Its height after $t$ seconds is modeled by:

$h(t) = -5t^2 + 20t$ (in metres)

For what time interval is the ball at least $15$ metres above the ground?

Solution Here, $-5t^2 + 20t \ge 15$

$-5t^2 + 20t - 15 \ge 0$

$5t^2 - 20t + 15 \le 0$

By solving the quadratic equation $5t^2 - 20t + 15 = 0$.

$t = \dfrac{-(-20) \pm \sqrt{(-20)^2 - 4(5)(15)}}{2(5)}$

$= \dfrac{20 \pm \sqrt{400 - 300}}{10}$

$= \dfrac{20 \pm \sqrt{100}}{10}$

$= \dfrac{20 \pm 10}{10}$

$= \dfrac{20 - 10}{10}, \dfrac{20 + 10}{10}$

$= 1, 3$

So, the ball is at least $15$ metres high when $1 \le t \le 3$ seconds.