Question

Starting from rest, a $2500\mathrm{kg}$helicopter accelerates straight up at a constant $2.0\mathrm{m}/{\mathrm{s}}^2.$What is the helicopter's height at the moment its blades are providing an upward force of $26\mathrm{kN}$? The helicopter can be modeled as a$2.6$-m-diameter sphere.

Short answer

The height of which a helicopter lifted by providing an upward force is$47265m.$

Step-by-step solution

Given

The mass of a helicopter is $2500\mathrm{kg}$and the acceleration of the helicopter straight up is $2.0\mathrm{m}/{\mathrm{s}}^2$. The upward force exerted by the helicopter is$26\mathrm{kN}$. The diameter of the sphere is $2.6\mathrm{m}$.

The weight of the helicopter act downward and the motion of the helicopter is in vertically upward. So, the drag force exerted by the air on the helicopter act vertically downward.

Formula used.

To calculate the lifting velocity of the helicopter using Newton's law is,

$\begin{array}{rl}& mg-F_{\mathrm{d}}=ma\\ & F_{\mathrm{d}}=mg+ma\left(1\right)\\ & F_{\mathrm{d}}=m(g+a)\end{array}$

• $m$is the mass of the helicopter.
• $F_{\mathrm{d}}$is the drag force.
• $a$is the constant acceleration.
• $g$is the acceleration due to gravity.

Formula to calculate the drag force on the helicopter by assuming it as a sphere is,

$F_{\mathrm{d}}=\frac{1}{2}\left(\frac{a{\mathrm{er}}^2}{4}\right)\rho v^2$

• $d$is the diameter of the sphere.
• $v$is the velocity of the helicopter uplifted.
• $\rho$is the density of air.

Calculation.

Substitute $\frac{1}{2}\left(\frac{\pi d^2}{4}\right)\rho v^2$for $F_{\mathrm{d}}$in equation (I).

$\begin{array}{c}\frac{1}{2}\left(\frac{\pi d^2}{4}\right)\rho v^2=m(g+a)\\ v^2=\frac{8m(g+a)}{\rho \mathrm{\Omega }{d}^2}\\ v=\sqrt{\frac{8m(g+a)}{\rho d{d}^2}}\end{array}$

The density of air is generally$1.2\mathrm{kg}/{\mathrm{m}}^3$

Substitute $2500\mathrm{kg}$for $m,9.8\mathrm{m}/{\mathrm{s}}^2$for $g,2.0\mathrm{m}/{\mathrm{s}}^2$for $a,1.2\mathrm{kg}/{\mathrm{m}}^3$for $\rho$and $2.6\mathrm{m}$for $d$to find $v$.

$v=\sqrt{\frac{8\left(2500\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2+2.0\mathrm{m}/{\mathrm{s}}^2\right)}{3.14\times (2.6\mathrm{m}{)}^2\left(1.2\mathrm{kg}/{\mathrm{m}}^2\right)}}$

$v=\sqrt{\frac{226000{\mathrm{m}}^2/{\mathrm{s}}^2}{25.47}}$

$=96.25\mathrm{m}/\mathrm{s}$.

Calculation.

Formula to calculate the vertical distance of the helicopter is,

$v^2=u^2+2gh$

• $u$is the initial velocity of the helicopter.
• $h$is the vertical height of the helicopter lifted.

Substitute $96.25\mathrm{m}/\mathrm{s}$for $v,0\mathrm{m}/\mathrm{s}$for $u$and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$to find $h$.

$\begin{array}{rl}(96.25\mathrm{m}/\mathrm{s}{)}^2& =(0\mathrm{m}/\mathrm{s}{)}^2+2(9.8\mathrm{m}/{\mathrm{s}}^2)h\\ h& =\frac{9204.06{\mathrm{m}}^2/{\mathrm{s}}^2}{19.6\mathrm{m}/{\mathrm{s}}^2}\\ & =472.65\mathrm{m}\end{array}$

Conclusion.

Therefore, the height of which a helicopter lifted by providing an upward force is$472.65m.$