Chapter 11.1, Problem 11.29P

The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as

$a=\frac{-32.2}{{\left[1+(y/20.9\times {10}^6)\right]}^2}$

where a and y are expressed in ft/s² and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.

Fig. P11.29

To determine

The height $\left(y_{\max }\right)$ reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is $1800\text{ft/s}$.

Answer to Problem 11.29P

The height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $1800\text{ft/s}$ is $\underset{\_}{50.4\times {10}^3\text{ft}}$.

Explanation of Solution

Given information:

The acceleration (a) due to gravity at an altitude y above the surface of the earth is $\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}$.

The initial velocity $\left(v_0\right)$ is $1800\text{ft/s}$.

Calculation:

Write the relation for the speed (a) as given below:

$a=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}$ (1)

Here, a is the acceleration and y is the altitude.

Express acceleration (a) by differentiation velocity (v) with respective to altitude (y):

$a=v\frac{dv}{dy}$

Substitute $\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}$ for a.

$\begin{array}{l}\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}=v\frac{dv}{dy}\\ vdv=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}dy\end{array}$

Apply integration.

${\displaystyle \underset{v_0}{\overset{0}{\int }}vdv}={\displaystyle \underset{0}{\overset{y_{\max }}{\int }}\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}dy}$

Integrate the equation.

$\begin{array}{l}{\left[\frac{v^2}{2}\right]}_{v_0}^0=-32.2{\left[20.9\times {10}^6\frac{1}{1+\frac{y}{20.9\times {10}^6}}\right]}_0^{y_{\max }}\\ 0-\frac{v_0^2}{2}=-32.2\left[20.9\times {10}^6\frac{1}{1+\frac{y_{\max }}{20.9\times {10}^6}}\right]-0\\ v_0^2=1345.96\times {10}^6\left[1-\frac{1}{1+\frac{y_{\max }}{20.9\times {10}^6}}\right]\end{array}$

Solve for $y_{\max }$.

$y_{\max }=\frac{v_0^2}{64.4-\frac{v_0^2}{20.9\times {10}^6}}$ (2)

Calculate the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $1800\text{ft/s}$.

Substitute $1800\text{ft/s}$ for t in Equation (3).

$\begin{array}{c}{y}_{\max }=\frac{{\left(1,800\right)}^2}{64.4-\frac{{\left(1,800\right)}^2}{20.9\times {10}^6}}\\ =\frac{3,240,000}{64.2449}\\ =50.4\times {10}^3\text{ft}\end{array}$

Therefore, the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $1800\text{ft/s}$ is $\underset{\_}{50.4\times {10}^3\text{ft}}$.

To determine

The height $\left(y_{\max }\right)$ reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is $3000\text{ft/s}$.

Answer to Problem 11.29P

The height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $3000\text{ft/s}$ is $\underset{\_}{140.7\times {10}^3\text{ft}}$.

Explanation of Solution

Given information:

The acceleration (a) due to gravity at an altitude y above the surface of the earth is $\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}$.

The initial velocity $\left(v_0\right)$ is $3000\text{ft/s}$.

Calculation:

Calculate the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $3000\text{ft/s}$.

Substitute $3,000\text{ft/s}$ for t in Equation (3).

$\begin{array}{c}{y}_{\max }=\frac{{\left(3,000\right)}^2}{64.4-\frac{{\left(3,000\right)}^2}{20.9\times {10}^6}}\\ =\frac{9,000,000}{63.969}\\ =140.7\times {10}^3\text{ft}\end{array}$

Therefore, the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $3000\text{ft/s}$ is $\underset{\_}{140.7\times {10}^3\text{ft}}$.

To determine

The height $\left(y_{\max }\right)$ reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is $3000\text{ft/s}$.

Answer to Problem 11.29P

The height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $36,700\text{ft/s}$ is $\underset{\_}{\infty }$.

Explanation of Solution

Given information:

The acceleration (a) due to gravity at an altitude y above the surface of the earth is $\frac{-32.2}{{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^2}$.

The initial velocity $\left(v_0\right)$ is $36,700\text{ft/s}$.

Calculation:

Calculate the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $36,700\text{ft/s}$.

Substitute $36,700\text{ft/s}$ for t in equation (3).

$\begin{array}{c}{y}_{\max }=\frac{{\left(36,700\right)}^2}{64.4-\frac{{\left(36,700\right)}^2}{20.9\times {10}^6}}\\ =\frac{1,346,890,000}{-0.0445}\\ =-3.03\times {10}^{10}\text{ft}\end{array}$

The above solution is invalid because the velocity does not reduce to zero.  The velocity (v) $36,700\text{ft/s}$ is beyond the escaped velocity $\left(v_R\right)$ from the earth. 

Therefore, the height $\left(y_{\max }\right)$ reached by a projectile if its initial velocity is $36,700\text{ft/s}$ is $\underset{\_}{\infty }$.