2022HSCShort answer

Q16 (b)

4 marks

Question

A projectile of mass $M\,\mathrm{kg}$ is launched vertically upwards from a horizontal plane with initial speed $v_0\,\mathrm{m\,s^{-1}}$ , where $v_0<100\,\mathrm{m\,s^{-1}}$ . The projectile experiences a resistive force of magnitude $0.1Mv$ newtons, where $v\,\mathrm{m\,s^{-1}}$ is the speed of the projectile. The acceleration due to gravity is $10\,\mathrm{m\,s^{-2}}$ .

The projectile lands on the horizontal plane $7$ seconds after launch.

Find the value of $v_0$ , correct to $1$ decimal place.

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	4
Obtains the velocity, or equivalent merit	3
Attempts to solve differential equation, or equivalent merit	2
Attempts to obtain the force equation	1

Sample answer:
$M\ddot y=-Mg-0.1Mv_y$
$\frac{dv_y}{dt}=-(g+0.1v_y)$

\int\frac{dv_y}{g+0.1v_y}=-\int dt

$10\left[\ln|g+0.1v_y|\right]=-t+k$
When $t=0$ , $v_y=v_0$ :
$10\ln|g+0.1v_0|=k$

\ln|g+0.1v_y|=-\frac t{10}+\ln|g+0.1v_0|

\ln\left|\frac{g+0.1v_y}{g+0.1v_0}\right|=-\frac t{10}

\ln\frac{g+0.1v_y}{g+0.1v_0}=-\frac t{10} \quad\text{as }g+0.1v>0\text{ because all speeds are less than }100\text{ m s}^{-1}

$g+0.1v_y=(g+0.1v_0)e^{-t/10}$
$0.1v_y=-g+(g+0.1v_0)e^{-t/10}$
$\therefore v_y=10\left[-g+(g+0.1v_0)e^{-t/10}\right]$

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Syllabus alignment

Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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