Chapter 1.5, Problem 111E

Falling-Body Problems Use the formula h = − 16t² + v₀t discussed in Example 9.

129. A ball is thrown straight upward at an initial speed of v₀ = 40 ft/s.

1. (a) When does the ball reach a height of 24 ft?
2. (b) When does it reach a height of 48 ft?
3. (c) What is the greatest height reached by the ball?
4. (d) When does the ball reach the highest point of its path?
5. (e) When does the ball hit the ground?

To determine

The time taken by the ball to reach a height of $24\text{ft}$ if the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$.

Answer to Problem 111E

The ball reaches a height of $24\text{ft}$ after $\underset{\_}{1\text{s}\text{and}\text{1}\frac{1}{2}\text{s}}$.

Explanation of Solution

Given:

An object thrown or fired straight upward at an initial speed of $v_0$ ft/s will reach a height of h feet after t seconds, where h and t are related by the formula $h=-16t^2+v_{0}t$.

Calculation:

The height of the ball is given as $24\text{ft}$ and the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$.

Therefore, substitute $24$ for $h$ and $40\text{ft}/\text{s}$ for $v_0$ in the equation $h=-16t^2+v_{0}t$ to obtain the time taken by the ball to reach a height of $24\text{ft}$.

$\begin{array}{l}24=-16t^2+40t\\ 16t^2-40t+24=0\end{array}$

Simplify further as follows.

$\begin{array}{l}8\left(2t^2-5t+3\right)=0\\ 2t^2-5t+3=0\\ 2t^2-2t-3t+3=0\\ 2t\left(t-1\right)-3\left(t-1\right)=0\end{array}$

On further simplifications, the following is obtained.

$\begin{array}{l}\left(t-1\right)\left(2t-3\right)=0\\ t-1=0\text{or }2t-3=0\\ t=1\text{or }t=\frac{3}{2}\\ t=1\text{or }t=1\frac{1}{2}\end{array}$

Thus, the ball reaches a height of $24\text{ft}$ after $\underset{\_}{1\text{s}\text{and}\text{1}\frac{1}{2}\text{s}}$.

To determine

The time taken by the ball to reach a height of $48\text{ft}$ if the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$.

Answer to Problem 111E

The ball $\underset{\_}{\text{never}}$ reaches the height of $48\text{ft}$.

Explanation of Solution

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

The height of the ball is given as $48\text{ft}$ and the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$.

Therefore, substitute $48$ for $h$ and $40\text{ft}/\text{s}$ for $v_0$ in the equation $h=-16t^2+v_{0}t$ to obtain the time taken by the ball to reach a height of $48\text{ft}$.

$\begin{array}{l}48=-16t^2+40t\\ 16t^2-40t+48=0\end{array}$

Simplify further as follows.

$\begin{array}{l}8\left(2t^2-5t+6\right)=0\\ 2t^2-5t+6=0\end{array}$

Compare this equation with the general form $a{x}^2+bx+c=0$.

Here $a=2,b=-5\text{and }c=6$.

Use the quadratic formula to obtain the roots of the given equation.

$\begin{array}{c}t=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\times 2\times 6}}{2\times 2}\\ t=\frac{5\pm \sqrt{25-48}}{4}\\ t=\frac{5\pm \sqrt{-23}}{4}\end{array}$

Since the square of any real number is nonnegative, $\sqrt{-23}$ is undefined in the real number system.

Thus, the ball $\underset{\_}{\text{never}}$ reaches the height of $48\text{ft}$.

To determine

The greatest height reached by the ball.

Answer to Problem 111E

The greatest height reached by the ball is $\underset{\_}{25\text{ft}}$.

Explanation of Solution

Result used:

Discriminant of a quadratic equation:

The quantity $b^2-4ac$ is called the discriminant of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$, from which the number of solutions the quadratic equation can have, can be obtained.

1. If $b^2-4ac>0$, there are two unequal real solutions.

2. If $b^2-4ac=0$, there is a repeated real solution, a double root.

3. If $b^2-4ac<0$, there is no real solution.

Calculation:

Given that the height of an object thrown straight upward at an initial speed of $v_0$ ft/s will reach a height of h feet after t seconds, where h and t are related by the formula $h=-16t^2+v_{0}t$.

Since the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$, substitute $v_0=40\text{ft}/\text{s}$.

$16t^2-40t+h=0$

Compare this equation with the general form $a{x}^2+bx+c=0$.

Here $a=2,b=-5\text{and }c=6$.

Note that, the ball will reach the highest point only once and the quadratic equation has only one solution when the discriminant is zero.

Therefore, let the discriminant be zero.

$\begin{array}{l}D=b^2-4ac\\ 0={\left(-40\right)}^2-4\times 16\times h\\ 0=1600-64h\\ 64h=1600\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}h=\frac{1600}{64}\\ h=25\end{array}$

Thus, the greatest height reached by the ball is $\underset{\_}{25\text{ft}}$.

To determine

The time taken by the ball to reach the highest point.

Answer to Problem 111E

The ball reaches the highest point after $\underset{\_}{1\frac{1}{4}\text{s}}$.

Explanation of Solution

The greatest height reached by the ball is obtained as $25\text{ft}$ and the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$,

Substitute $25$ for $h$ and $40$ for $v_0$ in the equation $h=-16t^2+v_{0}t$ to obtain the time taken by ball to reach the highest point.

$\begin{array}{l}25=-16t^2+40t\\ 16t^2-40t+25=0\end{array}$

Compare this equation with the general form $a{x}^2+bx+c=0$.

Here $a=16,b=-40\text{and }c=25$.

Use the quadratic formula to obtain the roots of the given equation.

$\begin{array}{c}t=\frac{-\left(-40\right)\pm \sqrt{{\left(-40\right)}^2-4\times 16\times 25}}{2\times 16}\\ t=\frac{40\pm \sqrt{1600-1600}}{32}\\ t=\frac{40}{32}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}t=\frac{5}{4}\\ t=1\frac{1}{4}\end{array}$

Thus, the ball reaches the highest point after $\underset{\_}{1\frac{1}{4}\text{s}}$.

To determine

The time taken by the ball to hit the ground.

Answer to Problem 111E

The ball hits the ground after $\underset{\_}{2\frac{1}{2}\text{s}}$.

Explanation of Solution

The height of a ball is $0$ when the ball hits the ground and the ball is thrown straight upward at an initial speed of $40\text{ft}/\text{s}$.

Substitute $0$ for $h$ and $40\text{ft}/\text{s}$ for $v_0$ in the equation $h=-16t^2+v_{0}t$.

$\begin{array}{l}0=-16t^2+40t\\ 16t^2-40t=0\\ t\left(16t-40\right)=0\\ t=0\text{or }16t-40=0\end{array}$

Simplify further as follows.

$\begin{array}{l}t=0\text{or }16t-40=0\\ t=0\text{or }16t=40\\ t=0\text{or }t=\frac{40}{16}\\ t=0\text{or }t=\frac{5}{2}\end{array}$

That is, $t=0\text{or }t=2\frac{1}{2}$.

Thus, the ball hits the ground after $\underset{\_}{2\frac{1}{2}\text{s}}$.