1.1 Q-6

Question Statement

A stone falls from a height of 60 m, and the height $h$ after $x$ seconds is approximately given by the equation:

h(x) = 40 - 10x^2

i. Find the height of the stone at the following times:

$x = 1$ sec
$x = 1.5$ sec
$x = 1.7$ sec

ii. When does the stone strike the ground?

Background and Explanation

This problem involves the motion of a falling object, with the height $h(x)$ given as a quadratic function of time $x$ . The function $h(x) = 40 - 10x^2$ describes the height of the stone at any given moment. The goal is to substitute the given time values into this equation and solve for the corresponding heights. Additionally, to find when the stone hits the ground, we need to solve for $x$ when the height becomes zero.

Solution

The height equation is:

h(x) = 40 - 10x^2

i. Find the height at different times.

a) When $x = 1$ sec:

Substitute $x = 1$ into the equation:

h(1) = 40 - 10(1)^2 = 40 - 10 = 30 , \text{m}

So, the height at $x = 1$ sec is 30 m.

b) When $x = 1.5$ sec:

Substitute $x = 1.5$ into the equation:

h(1.5) = 40 - 10(1.5)^2 = 40 - 10(2.25) = 40 - 22.5 = 17.5 , \text{m}

So, the height at $x = 1.5$ sec is 17.5 m.

c) When $x = 1.7$ sec:

Substitute $x = 1.7$ into the equation:

h(1.7) = 40 - 10(1.7)^2 = 40 - 10(2.89) = 40 - 28.9 = 11.1 , \text{m}

So, the height at $x = 1.7$ sec is 11.1 m.

ii. When does the stone strike the ground?

The stone strikes the ground when $h(x) = 0$ . Set the height equation equal to zero and solve for $x$ :

0 = 40 - 10x^2

Rearrange:

10x^2 = 40

x^2 = 4

x = 2 , \text{sec}

So, the stone strikes the ground after 2 seconds.

Key Formulas or Methods Used

The height of the stone at any time is given by the quadratic function $h(x) = 40 - 10x^2$ .
To find the height at a given time, substitute the time into the function.
To find when the stone strikes the ground, set $h(x) = 0$ and solve for $x$ .

Summary of Steps

Use the given height equation $h(x) = 40 - 10x^2$ to calculate the height at different times.
For $x = 1$ , $x = 1.5$ , and $x = 1.7$ , substitute these values into the equation to find the corresponding heights.
To determine when the stone strikes the ground, set $h(x) = 0$ and solve for $x$ . The stone hits the ground after 2 seconds.