The terminal velocity $v_t$ of a skydiver is given by the relationship $$v_t = \frac{2mg}{\sqrt{\rho AC_d}}$$ where - $m$ is the mass of the skydiver in kg - $g$ is the gravitational field strength in N kg$^{-1}$ - $C_d$ is the drag coefficient - $\rho$ is the density of air in kg m$^{-3}$ - $A$ is the area of the skydiver in m$^2$. When in freefall, a skydiver of mass 95 kg has a drag coefficient of 1.0 and a terminal velocity of 44 m s$^{-1}$. The gravitational field strength is 9.8 N kg$^{-1}$ and the density of air is 1.21 kg m$^{-3}$. The area of the skydiver is A 0-59 m$^2$ B 0-79 m$^2$ C 0-89 m$^2$ D 4-2 m$^2$ E 35 m$^2$

Scottish Highers Physics Unseen Formula/Graph Plotting: The terminal velocity $v

The terminal velocity $v_t$ of a skydiver is given by the relationship $$v_t = \frac{2mg}{\sqrt{\rho AC_d}}$$ where - $m$ is the mass of the skydiver in kg - $g$ is the gravitational field strength in N kg$^{-1}$ - $C_d$ is the drag coefficient - $\rho$ is the density of air in kg m$^{-3}$ - $A$ is the area of the skydiver in m$^2$ - Scottish Highers Physics - Question 7 - 2019

Question 7

$The-terminal-velocity-$v_t$-of-a-skydiver-is-given-by-the-relationship--$$v_t-=-\frac{2mg}{\sqrt{\rho-AC_d}}$$--where----$m$-is-the-mass-of-the-skydiver-in-kg---$g$-is-the-gravitational-field-strength-in-N-kg$^{-1}$---$C_d$-is-the-drag-coefficient---$\rho$-is-the-density-of-air-in-kg-m$^{-3}$---$A$-is-the-area-of-the-skydiver-in-m$^2$-Scottish Highers Physics-Question 7-2019.png$

The terminal velocity $v_t$ of a skydiver is given by the relationship $$v_t = \frac{2mg}{\sqrt{\rho AC_d}}$$ where - $m$ is the mass of the skydiver in kg - $g$ ... show full transcript

Worked Solution & Example Answer:The terminal velocity $v_t$ of a skydiver is given by the relationship $$v_t = \frac{2mg}{\sqrt{\rho AC_d}}$$ where - $m$ is the mass of the skydiver in kg - $g$ is the gravitational field strength in N kg$^{-1}$ - $C_d$ is the drag coefficient - $\rho$ is the density of air in kg m$^{-3}$ - $A$ is the area of the skydiver in m$^2$ - Scottish Highers Physics - Question 7 - 2019

Step 1

Calculate the required area of the skydiver:

96%

114 rated

Answer

To find the area $A$ , rearrange the terminal velocity equation:

$A = \frac{2mg}{\rho C_d v_t^2}$

Substituting the values:

$m = 95$ kg
$g = 9.8$ N kg $^{-1}$
$C_d = 1.0$
$\rho = 1.21$ kg m $^{-3}$
$v_t = 44$ m s $^{-1}$

This gives:

$A = \frac{2 \times 95 \times 9.8}{1.21 \times 1.0 \times (44)^2}$

Calculating this:

Compute the numerator: $2 \times 95 \times 9.8 = 1864$ .
Compute the denominator: $1.21 \times 1.0 \times 1936 = 2343.52$ .
Now divide: $A = \frac{1864}{2343.52} \approx 0.796$ m $^2$ .

Therefore, the closest option is B: 0-79 m $^2$ .

Calculate the required area of the skydiver:

Join the Scottish Highers students using SimpleStudy...