Early test flights for the space shuttle used a “glider” (mass of 980 kg including pilot). After a horizontal launch at at a height of 3500 m, the glider eventually landed at a speed of (a) What would its landing speed have been in the absence of air resistance? (b) What was the average force of air resistance exerted on it if it came in at a constant glide angle of to the Earth’s surface?

If the total work done by all external and internal forces is zero, then the mechanical energy remains constant.

It is given by:

\(P{E_{\rm{i}}} + K{E_{\rm{i}}} = P{E_{\rm{f}}} + K{E_{\rm{f}}}\)

Here, \(P{E_{\rm{i}}}\) is the initial potential energy, \(K{E_{\rm{i}}}\) is the initial kinetic energy, \(P{E_{\rm{f}}}\) is the final potential energy, and \(K{E_{\rm{f}}}\) is the final kinetic energy of the system.

The combined mass of the glider and the pilot is\(m = 980\;{\rm{kg}}\).

The launch speed of the glider is\(u = 480\;{\rm{km/h}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right) = 133.3\;{\rm{m/s}}\).

The final/landing speed of the glider is

\(v = 210\;{\rm{km/h}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right) = 58.3\;{\rm{m/s}}\).

The initial height attained is \({h_{\rm{i}}} = 3500\;{\rm{m}}\).

The final height after landing is\({h_{\rm{f}}} = 0\).

Apply the law of conservation of mechanical energy at a height and at the ground.

\(\begin{aligned}mg{h_{\rm{i}}} + \frac{1}{2}mv_{\rm{i}}^2 &= mg{h_{\rm{f}}} + \frac{1}{2}mv_{\rm{f}}^2\\mg{h_{\rm{i}}} + \frac{1}{2}mv_{\rm{i}}^2 &= 0 + \frac{1}{2}mv_{\rm{f}}^2\\v_{\rm{f}}^2 &= 2g{h_{\rm{i}}} + v_{\rm{i}}^2\\{v_{\rm{f}}} &= \sqrt {2g{h_{\rm{i}}} + v_{\rm{i}}^2} \end{aligned}\) … (i)

Substitute the known values in equation (i).

\(\begin{aligned}{v_{\rm{f}}} &= \sqrt {2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {3500\;{\rm{m}}} \right) + {{\left( {133.3\;{\rm{m/s}}} \right)}^2}} \\ &= 293.88\;{\rm{m/s}}\\ &\approx 294\;{\rm{m/s}}\end{aligned}\)

Thus, the landing speed of the gliderin the absence of air resistance is \(294\;{\rm{m/s}}\).

The work-energy theorem defines the relation between work done and kinetic energy. It states that the net work done on the object is equal to the change in kinetic energy. It can be written as:

\(W = \Delta KE\)

The air resistance does negative work on the glider and reduces the final speed of the glider from its theoretical value determined in part (a). The net work done is equal to the difference between the kinetic energy in the absence of air resistance and the actual kinetic energy upon landing. Applying the work-energy theorem, you get:

\(\begin{aligned}W &= KE - K{E_{{\rm{air resistance}}}}\\ &= \frac{1}{2}mv_{\rm{f}}^2 - \frac{1}{2}m{v^2}\\ &= \frac{1}{2}\left( {980\;{\rm{kg}}} \right)\left[ {{{\left( {294\;{\rm{m/s}}} \right)}^2} - {{\left( {58.3\;{\rm{m/s}}} \right)}^2}} \right]\\ &= 4.069 \times {10^7}\;{\rm{J}}\end{aligned}\)

\(\begin{aligned}d &= \frac{{3500\;{\rm{m}}}}{{\sin \left( {{{12}^{\rm{o}}}} \right)}}\\ &= 16834.07\;{\rm{m}}\end{aligned}\)

The work done is given by:

\(\begin{aligned}W &= F \cdot d\\F &= \frac{W}{d}\\ &= \frac{{4.069 \times {{10}^7}\;{\rm{J}}}}{{16834.07\;{\rm{m}}}}\\ &= 2417.12\;{\rm{N}}\end{aligned}\)

Thus, the magnitude of the force of air resistance is2417.12 N.