Chapter 12.1, Problem 12.25P

To determine

(a)

The maximum height of the trajectory.

Answer to Problem 12.25P

The maximum height of the trajectory: $h=335.3m$

Explanation of Solution

Given information:

Mass of the projectile: $m=4kg$

Initial velocity: $v_0=90m/s$

The aerodynamic drag: $D=\left(0.0024\right)v^2$

Direction of the aerodynamic drag is always opposite to the direction of the velocity.

Calculations:

Consider the motion of the projectile from the ground to the maximum height. The free body diagram and kinetic diagram are drawn as shown below.

Applying Newton’s second law for the vertical motion:

$\sum F_v=ma;$

$-D-mg=ma$

$or,$

$a=\frac{-D-mg}{m}$

$Substitut\text{i}\text{n}g,a=v\frac{dv}{dx}\text{ and }D=\left(0.0024\right)v^2$

$v\frac{dv}{dx}=\frac{-\left(0.0024\right)v^2-mg}{m}$

$\text{Rewriting the above equation:}$

$\frac{vdv}{mg+0.0024v^2}=-\frac{dx}{m}\mathrm{.................}\left(1\right)$

$\text{Since the projectile move from the ground to the maximum height}$

$\therefore \text{i}\text{n}itialheight=0,\text{i}\text{n}itialvelocity=90m/s$

$f\text{i}\text{n}alheight=h,f\text{i}\text{n}alvelocity=0$

$\text{Hence, integrating eq}\text{.}\left(\text{1}\right)\text{ with the known limits:}$

${\int }_{90}^0\frac{vdv}{mg+0.0024v^2}={\int }_0^h-\frac{dx}{m}$

$\frac{1}{0.0024}{\int }_{90}^0\frac{vdv}{\frac{mg}{0.0024}+v^2}={\int }_0^h-\frac{dx}{m}$

$\frac{1}{0.0024}{\left[\frac{1}{2}\ln \left(\frac{mg}{0.0024}+v^2\right)\right]}_{90}^0=-\frac{1}{m}{\left[x\right]}_0^h$

$\frac{1}{0.0024}\left[\frac{1}{2}\ln \left(\frac{mg}{0.0024}+0\right)-\frac{1}{2}\ln \left(\frac{mg}{0.0024}+{90}^2\right)\right]=-\frac{h}{m}$

$\frac{1}{0.0024}\left[\frac{1}{2}\ln \left(\frac{mg}{mg+19.44}\right)\right]=-\frac{h}{m}$

$or,$

$h=\frac{-m}{0.0048}\left[\ln \left(\frac{mg}{mg+19.44}\right)\right]$

$Substitut\text{i}\text{n}g,m=4kgandg=9.81m/s^2$

$h=\frac{-4}{0.0048}\left[\ln \left(\frac{4\left(9.81\right)}{4\left(9.81\right)+19.44}\right)\right]$

$\Rightarrow h=335.3m$

Conclusion:

The maximum height of the trajectory is $h=335.3m$

To determine

(b)

The speed of the projectile when it reaches the ground.

Answer to Problem 12.25P

The speed of the projectile: $v=73.6m/s$

Explanation of Solution

Given information:

Mass of the projectile: $m=4kg$

Initial velocity: $v_0=90m/s$

The maximum height of the trajectory: $h=335.3m$

The aerodynamic drag: $D=\left(0.0024\right)v^2$

Direction of the aerodynamic drag is always opposite to the direction of the velocity.

Calculations:

Now, consider the motion of the projectile from the maximum height to the ground. The free body diagram and the kinetic diagram are drawn as shown below.

Applying Newton’s second law for the downward motion:

$\sum F_v=-ma;$

$D-mg=-ma$

$or,$

$a=\frac{mg-D}{m}$

$Substitut\text{i}\text{n}g,a=v\frac{dv}{dx}\text{ and }D=\left(0.0024\right)v^2$

$v\frac{dv}{dx}=\frac{mg-\left(0.0024\right)v^2}{m}$

$\text{Rewriting the above equation:}$

$\frac{vdv}{mg-0.0024v^2}=\frac{dx}{m}\mathrm{.................}\left(2\right)$

$\text{Since the projectile move from the maximum height to the ground}\text{.}$

$\therefore \text{i}\text{n}itialheight=h=335.3m,\text{i}\text{n}itialvelocity=0m/s$

$f\text{i}\text{n}alheight=0,f\text{i}\text{n}alvelocity=v$

$\text{Hence, integrating eq}\text{.}\left(2\right)\text{ with the known limits:}$

${\int }_0^v\frac{vdv}{mg-0.0024v^2}={\int }_h^0\frac{dx}{m}$

$\frac{1}{0.0024}{\int }_0^v\frac{vdv}{\frac{mg}{0.0024}-v^2}={\int }_{-335.3}^0\frac{dx}{m}$

$\frac{1}{0.0024}{\left[-\frac{1}{2}\ln \left(1-\frac{0.0024v^2}{mg}\right)\right]}_0^v=\frac{1}{m}{\left[x\right]}_{-335.3}^0$

$\frac{-1}{0.0048}\ln \left(\frac{mg-0.0024v^2}{mg}\right)=\frac{335.3}{m}$

$or,$

$mg-0.0024v^2=mg{e}^{\frac{-1.60961}{m}}$

$\therefore v=\sqrt{\frac{mg\left(1-e^{\frac{-1.60961}{m}}\right)}{0.0024}}$

$Substitut\text{i}\text{n}g,m=4kgandg=9.81m/s^2$

$v=\sqrt{\frac{4\times 9.81\left(1-e^{\frac{-1.60961}{4}}\right)}{0.0024}}$

$\Rightarrow v=73.6m/s$

Conclusion:

The speed of the projectile when it reaches the ground is $v=73.6m/s$.