Question

The "Vomit Comet." In microgravity astronaut training and equipment testing, NASA flies a $\mathbf{KC}\mathbf{135}\mathbf{A}$aircraft along a parabolic flight path. As shown in Figure, the aircraft climbs from $\mathbf{24000}\mathbf{}\mathbf{ft}$to$\mathbf{31000}\mathbf{}\mathbf{ft}$, where it enters a parabolic path with a velocity of $\mathbf{143}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$nose high at ${\mathbf{45}}^{\mathbf{\circ }}$and exits with velocity $\mathbf{143}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at ${\mathbf{45}}^{\mathbf{\circ }}$nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free fall; astronauts and equipment float freely as if there were no gravity. What are the aircraft's

(a) speed and

(b) altitude at the top of the maneuver?

(c) What is the time interval spent in microgravity?

Short answer

(a) The speed of the aircraft is$101\mathrm{m}/\mathrm{s}$.

(b) The altitude at the top of the maneuver is$32.7\times {10}^3\mathrm{ft}$.

(c) The time interval spent in microgravity is $20.6\mathrm{s}$.

Step-by-step solution

Given data

Given info: The velocity of the aircraft when enters the parabolic path is $143\mathrm{m}/\mathrm{s}$with high nose at localid="1663604763353" ${45}^{\circ }$, velocity of the aircraft when exits the parabolic path is $143\mathrm{m}/\mathrm{s}$with nose low at ${45}^{\circ }$

The aircraft climbs from $24000\text{ft}$to $31000\text{ft}$.

The speed of the aircraft.

From the Figure the aircraft speed at the top is given by,

${\mathit{v}}_{\mathbf{x}}\mathbf{=}{\mathit{v}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$ ...... (1)

Here,

${\mathit{v}}_{\mathbf{1}}$is the velocity of the aircraft.

$\mathit{\theta }$is the angle made by the aircraft with the horizontal.

(a) Determine the speed of the aircraft.

Substitute$143\mathrm{m}/\mathrm{s}$for$v_1$and${45}^{\circ }$for$\theta$in the equation (1),

role="math" localid="1663605058637" $\begin{array}{c}{v}_{\mathrm{x}}=(143\mathrm{m}/\mathrm{s})\cos \left(45°\right)\\ =101.116\mathrm{m}/\mathrm{s}\\ \approx 101\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the speed of the aircraft is $101\mathrm{m}/\mathrm{s}$.

(b) Determine the altitude at the top of the maneuver.

The formula to calculate the altitude at the top of the maneuver is,

${\left({\left(v_2\right)}_{\mathrm{y}}\right)}^2={\left({\left(v_1\right)}_{\mathrm{y}}\right)}^2+2a\left(y_2-y_1\right)$...... (2)

Here,

${\left(v_2\right)}_{\mathrm{y}}$is the initial speed of the aircraft in the $y$direction at the top of the maneuver.

is the acceleration of the aircraft.

${\left(v_1\right)}_{\mathrm{y}}$is the final speed of the aircraft in the $y$direction at the top of the maneuver.

$y_1$is the heights altitude.

$y_2$is the altitude at the top of the maneuver.

From the Figure the initial speed of the aircraft in the $y$direction at the top of the maneuver is 0,${\left(v_2\right)}_{\mathrm{y}}=0$.

The aircraft's initial speed at the top in direction is 0,role="math" localid="1663605342623" ${\left(v_2\right)}_{\mathrm{y}}=0$

${\left(v_1\right)}_{\mathrm{y}}=v_1\sin \theta$

Substitute$143\mathrm{m}/\mathrm{s}$for$v_1$and${45}^{\circ }$for$\theta$in the above equation.

${\left(v_1\right)}_{\mathrm{y}}=(143\mathrm{m}/\mathrm{s})\sin \left({45}^{\circ }\right)$

Substitute 0 for${\left(v_2\right)}_{\mathrm{y}},(143\mathrm{m}/\mathrm{s})\sin \left({45}^{\circ }\right)$for${\left(v_1\right)}_{\mathrm{y}},9.8\mathrm{m}/{\mathrm{s}}^2$for$g$, and$31000\text{ft}$for$y_1$in equation (2),

$\begin{array}{c}0={\left((143\mathrm{m}/\mathrm{s})\sin \left(45\right)\right)}^2+2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(y_2-31000\mathrm{ft}\right)\\ {\left((143\mathrm{m}/\mathrm{s})\sin \left(45\right)\right)}^2=2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(y_2-31000\mathrm{ft}\right)\\ \left(y_2-31000\mathrm{ft}\right)=\frac{{\left((143\mathrm{m}/\mathrm{s})\sin \left(45\right)\right)}^2}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ \left(y_2-31000\mathrm{ft}\right)=521.66\mathrm{m}\end{array}$

Further solve the above equation.

role="math" localid="1663605730322" $\begin{array}{c}\left(y_2-31000\mathrm{ft}\right)=521.66\times \frac{1}{0.3048} \text{ft}\\ y_2=32.7\times {10}^3 \text{ft}\end{array}$

Therefore, the altitude at the top of the maneuver is $32.7\times {10}^3 \text{ft}$.

(c) The time interval spent in microgravity.

From the free fall motion the equation is,

${\left(v_2\right)}_{\mathrm{y}}={\left(v_1\right)}_{\mathrm{y}}+gt$

Here,

$g$is the acceleration due to gravity.

From part (a) speed of the aircraft is$101\mathrm{m}/\mathrm{s}$.

Substitute$-101\mathrm{m}/\mathrm{s}$for${\left(v_2\right)}_{\mathrm{y}},101\mathrm{m}/\mathrm{s}$for${\left(v_1\right)}_y$and$9.8\mathrm{m}/{\mathrm{s}}^2$for$g$in the above equation.

$\begin{array}{c}-101\mathrm{m}/\mathrm{s}=101\mathrm{m}/\mathrm{s}+\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t\\ 202\mathrm{m}/\mathrm{s}=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t\\ t=\frac{202\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =20.6\mathrm{s}\end{array}$

Therefore, the time interval spent in microgravity is $26.0\text{s}$.