Chapter 3.3, Problem 74AYU

Analyzing the Motion of a Projectile A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal, with a muzzle velocity of $100$ feet per second. The height $h$ of the projectile is modeled by

$h(x)=\frac{-32x^2}{{100}^2}+x$

Where $x$ is the horizontal distance of the projectile from the firing point.

At what horizontal distance from the firing point is the height of the projectile

Find the maximum height of the projectile.

At what horizontal distance from the firing point will the projectile strike the ground?

Graph the function $h$ $0\le x\le 350$.

Use a graphing utility to verify the results obtained in parts $(b)$ and $(c)$.

When the height of the projectile is $50$ feet above the ground, how far has it travelled horizontally?

To determine

To calculate: The horizontal distance from the firing point at which the height of the projectile is maximum.

Answer to Problem 74AYU

Solution:

The height of projectile is maximum at a distance $156.25feet$ from the firing point

Explanation of Solution

Given Information:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point

Formula used:

The graph of the function $f(x)=a{x}^2+bx+c$ is a parabola and it has the maximum height at vertex given as $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$

Calculation:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point

The graph of the function $f(x)=a{x}^2+bx+c$ is a parabola and it has the maximum height at vertex given as $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$

Compare $h(x)=\frac{-32x^2}{{100}^2}+x$ with $f(x)=a{x}^2+bx+c$

$a=\frac{-32}{{100}^2},b=1,c=0$

$\frac{-b}{2a}=\frac{-1}{2\left(\frac{-32}{{100}^2}\right)}=\frac{{100}^2}{64}=156.25$

The height of projectile is maximum at a distance $156.25feet$ from the firing point

To determine

To calculate: The maximum height of the projectile, if the height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point.

Answer to Problem 74AYU

Solution:

The maximum height of the projectile is $78.125feet$

Explanation of Solution

Given Information:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point

Formula used:

The graph of the function $f(x)=a{x}^2+bx+c$ is a parabola and it has the maximum height at vertex given as $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$

Calculation:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point.

The graph of the function $f(x)=a{x}^2+bx+c$ is a parabola and it has the maximum height at vertex given as $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$

Compare $h(x)=\frac{-32x^2}{{100}^2}+x$ with $f(x)=a{x}^2+bx+c$

$a=\frac{-32}{{100}^2},b=1,c=0$

$\frac{-b}{2a}=\frac{-1}{2\left(\frac{-32}{{100}^2}\right)}=\frac{{100}^2}{64}=156.25$

$f\left(\frac{-b}{2a}\right)=f(156.25)$

$f(156.25)=\frac{-32}{{100}^2}{\left(156.25\right)}^2+(156.25)$

$f(156.25)=78.125$

The maximum height of projectile is $78.125feet$.

To determine

To calculate: The horizontal distance from the firing point at which the projectile will strike the ground, if the height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$, where $x$ is the horizontal distance of projectile from the firing point.

Answer to Problem 74AYU

Solution:

The projectile will strike the ground at $312.5feet$ from the firing point.

Explanation of Solution

Given Information:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$

Where $x$ is the horizontal distance of projectile from the firing point

Formula used:

When the projectile touches the ground the height of the projectile becomes zero.

$\Rightarrow h(x)=0$

Calculation:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$

Where $x$ is the horizontal distance of projectile from the firing point.

When the projectile touches the ground the height of the projectile becomes zero.

$\Rightarrow h(x)=0$

$\frac{-32x^2}{{100}^2}+x=0$

The least common denominator is ${100}^2$

$\frac{-32x^2+{100}^{2}x}{{100}^2}=0$

$-32x^2+10000x=0$

Divide by $-4$ on both sides

$8x^2-2500x=0$

The equation $8x^2-2500x=0$ is solved by using quadratic formula as

$x=\frac{-(2500)\pm \sqrt{{(-2500)}^2-4(8)(0)}}{2(8)}$

$x=\frac{2500\pm \sqrt{6250000}}{16}$

$x=\frac{2500\pm 2500}{16}$

$x=\frac{2500+2500}{16}$ or $x=\frac{2500-2500}{16}$

$x=312.5$ or $x=0$

As $x$ is the horizontal distance of projectile from the face of the cliff, so $x$ is always positive and greater than zero.

$\Rightarrow x=312.5$

The projectile will strike the ground at $312.5feet$ from the firing point.

To determine

To graph: To graph $h(x)=\frac{-32x^2}{{100}^2}+x,0\le x\le 350$

Explanation of Solution

Given Information:

$h(x)=\frac{-32x^2}{{100}^2}+x,0\le x\le 350$

Graph:

The equation of $h$ is $h(x)=\frac{-32x^2}{{100}^2}+x,0\le x\le 350$

To graph using graphing utility, follow the steps given below

Step 1: Press Y= and enter the expression $\frac{-32x^2}{{100}^2}+x$ in the numerator and enter $(0\le x)and(x\le 350)$ in the denominator.

Press [2^nd][MATH] to write the inequality

Press [2^nd][MATH][$⊳$] to get “and”

Step 2: Press [GRAPH].

The graph will be as given below:

Interpretation:

The graph of the equation $h(x)=\frac{-32x^2}{{100}^2}+x,0\le x\le 350$ will look like a parabola given above.

To determine

To calculate: The maximum and zeros of the function $h(x)=\frac{-32x^2}{{100}^2}+x$

Answer to Problem 74AYU

Solution:

Maximum of $h(x)=\frac{-32x^2}{{100}^2}+x$ is $78.125$ and zero of $h(x)=\frac{-32x^2}{{100}^2}+x$ is at $312.5$

Explanation of Solution

Given Information:

$h(x)=\frac{-32x^2}{{100}^2}+x$

Calculation:

The equation of $h$ is $h(x)=\frac{-32x^2}{{100}^2}+x$

To find the zeros and maximum of $h$ using graphing utility follow the steps given below

Step 1: Press Y= and enter the expression $h(x)=\frac{-32x^2}{{100}^2}+x$

Step 2: Press [2^nd][TRACE] to get the CALC. Under CALC menu choose 2: zero and press

ENTER.

Step 3: Move the cursor( use arrow keys ) to left of observed zero location. Press ENTER.

Step 4: Move the cursor( use arrow keys ) to right of observed zero location. Press ENTER.

Step 5: When the last screen asks for guess, press ENTER.

The coordinates of zeros are $\left(0,0\right),(312.5,0)$.

As the distance and height should be positive, the co-ordinate of zero is $(312.5,0)$

Step 6: Press [2^nd][TRACE] to get the CALC. Under CALC menu choose 4: maximum and press

ENTER.

Step 7: Move the cursor( use arrow keys ) to left of observed maximum location. Press ENTER.

Step 8: Move the cursor( use arrow keys ) to right of observed maximum location. Press ENTER.

Step 9: When the last screen asks for guess, press ENTER

The coordinate of maximum value is $(156.25,78.125)$

From part (b) the maximum height is $78.125$ and from part (c) the zero of the height function is $312.5$.

Hence, using graphical utility the maximum height and the zero of the function coincides with the answers from part (b) and part (c).

To determine

To calculate: The distance of projectile from the firing point, when it is 50 feet above the ground.

Answer to Problem 74AYU

Solution:

The distance of projectile from the firing point is $62.5feet$, when it is 50 feet above the ground.

Explanation of Solution

Given Information:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$

Where $x$ is the horizontal distance of projectile from the firing point

Formula used:

$h(x)=\frac{-32x^2}{{100}^2}+x$

Calculation:

A projectile is fired at an inclination of ${45}^{\circ }$ to the horizontal with a muzzle velocity of 100 feet per second. The height $h$ of the projectile is modelled by $h(x)=\frac{-32x^2}{{100}^2}+x$

Where $x$ is the horizontal distance of projectile from the firing point

The projectile is 50 feet above the ground.

$\Rightarrow h(x)=50$

$\frac{-32x^2}{{100}^2}+x=50$

The least common denominator is ${100}^2$

$\frac{-32x^2+{100}^{2}x}{{100}^2}=\frac{50{(100)}^2}{{100}^2}$

$-32x^2+10000x=500000$

Divide by $-4$ on both sides

$8x^2-2500x=-125000$

Add $125000$ on both sides

$8x^2-2500x+125000=0$

The equation $8x^2-2500x+125000=0$ is solved by using quadratic formula as

$x=\frac{-(2500)\pm \sqrt{{(-2500)}^2-4(8)(125000)}}{2(8)}$

$x=\frac{2500\pm \sqrt{6250000-4000000}}{16}$

$x=\frac{2500\pm \sqrt{2250000}}{16}$

$x=\frac{2500\pm 1500}{16}$

$x=\frac{2500+1500}{16}$ or $x=\frac{2500-1500}{16}$

$x=250$ or $x=62.5$

The height of projectile is maximum at a distance $156.25feet$ from the firing point.

So, when the projectile is at height 50 feet above the ground the distance from firing point must be less than 156.25 feet.

$\Rightarrow x=62.5$

The distance of projectile from the firing point is $62.5feet$, when it is 50 feet above the ground.