Question

A helicopter carrying dr. evil takes off with a constant upward acceleration of 5.0\ m/s^2. Secret agent austin powers jumps on just as the helicopter

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asked 2021-03-26
A helicopter carrying dr. evil takes off with a constant upward acceleration of \(\displaystyle{5.0}\ \frac{{m}}{{s}^{{2}}}\). Secret agent austin powers jumps on just as the helicopter lifts off the ground. Afterthe two men struggle for 10.0 s, powers shuts off the engineand steps out of the helicopter. Assume that the helicopter is infree fall after its engine is shut off and ignore effects of airresistance.
a) What is the max height above ground reached by the helicopter?
b) Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 m/s2. how far is powers above the ground when the helicopter crashes into the ground.

Expert Answers (2)

2021-03-28
Continuation of solution. Previous post provided part of solution for helicopter. Now, we also need to determine whenthe helicopter will hit the ground. Recall that the height ofthe helicopter is given by:
\(\displaystyle{y}{\left({t}\right)}={50}\cdot{t}^{{2}}-{4.9}\cdot{{t}_{{2}}^{{2}}}+{250}\)
where "t_2" is used as a reminder that the equation is valid for t>=10 (i.e. interval 2).
The helicopter reaches the ground when y(t) = 0. Note that this is a quadratic equation and therefore has two solutions. In this case, one solution is negative and the other is positive. The equation is only valid for \(\displaystyle{t}_{{2}}\geq{0}\), so we take the positive solution. I'll leave the details of solving y(t)=0 for you. The result is:
\(\displaystyle{t}_{{{2}{e}{n}{d}}}={13.88}\ {\sec{{o}}}{n}{d}{s}\)
DON'T FORGET. The time above is referenced to interval2. That interval started at time t=10 seconds. Therefore the total time that the helicopter was in the airs:
13.88 + 10 = 14.88 seconds.
AUSTIN POWERS
Now, what happened to Austin? He left the helicopter att=10 seconds (or t2=0, for interval 2). At that point he hadthe same velocity as the helicopter and was experiencing "freefall" (per the problem statement). Austin experiences freefall for 7 seconds (interval 2 time = 7 seconds). Since theonly influence on his motion is that due to acceleration ofgravity, then the equation for y(t) determined for the helicopteris also applicable to Austin:
\(\displaystyle{y}{\left({t}\right)}={50}\cdot{t}_{{2}}-{4.9}\cdot{{t}_{{2}}^{{2}}}+{250}\)
and
\(\displaystyle{y}{\left({7}\right)}={359.9}\) meters
Similary, the velocity equation determined before also appliesto Austin (again due to "free fall" assumption)
\(\displaystyle{v}{\left({t}\right)}={50}-{9.8}\cdot{{t}_{{2}}^{{2}}}\)
\(\displaystyle{v}{\left({7}\right)}=-{18.6}\ \frac{{m}}{{s}}\)
Now, we needed that information because Austin's motion will change at \(\displaystyle{t}_{{2}}={7}\) seconds. As stated before, the velocity and height at \(\displaystyle{t}_{{2}}={7}\) seconds will serve as initial conditions for the equations we need to determine that are applicable for \(\displaystyle{t}_{{2}}\) after 7 seconds. We will treat this as interval 3, and it applies to Austinonly.
According to the problem, during interval 3, Austin has anacceration of \(\displaystyle-{2}\frac{{m}}{{s}^{{2}}}\). We determined his velocity and position at that time. So the new equations of motion for Austin are;
\(\displaystyle{v}{\left({t}\right)}=-{2}\cdot{t}_{{3}}+{C}\quad{v}{\left({0}\right)}=-{18.6}\ \frac{{m}}{{s}}\), therefore C=-18.6 m/s
\(\displaystyle{v}{\left({t}\right)}=-{2}\cdot{t}_{{3}}-{18.6}\)
and his height is given by the integral of the velocity equation:
\(\displaystyle{y}{\left({t}\right)}=-{{t}_{{3}}^{{2}}}-{18.6}{t}_{{3}}+{C}\)
using the initial condition y(0)=359.9 meters, then C=359.9
\(\displaystyle{y}{\left({t}\right)}=-{{t}_{{3}}^{{2}}}-{18.6}\cdot{t}_{{3}}+{359.9}\)
Now we want to know Austin's height when the helicopter hits the ground. We calculated this as \(\displaystyle{t}_{{2}}={13.88}\) seconds. However, this is the time from when the helicopter (and Austin) began to experience free fall. Austin's motion changed 7 seconds after that and our y(t) equation for Austin is only validafter the 7 seconds. So we evaluate y(t) for Austin at time:
\(\displaystyle{t}_{{3}}={13.88}-{7}={6.88}\) seconds
Making the proper substitution:
y(6.88) = 184.59 meters
That is, Austin is still 184.59 meters above the ground when thehelicopter hits the ground! You should check my math forerrors.
A note about the integrations. We integrate acceleration toget velocity and integrate velocity to get position. Or, wecould state that acceleration is the rate of change of velocity,and velocity is the rate of change of position. Associatedwith each integration was a "constant of integration" that Idenoted as "C". You must always include this when determiningan "indefinite' integral (one that does not have upper and lowerlimits of integration). If we integrate an equation, and thatintegration includes a constant "C" then what happens if wedifferentiate our solution? The derivative of a constant is0! So, the "C" assures a general solution to ourintegration. For this physics problem you could look at the"contant of integration" providing information as to the state ofthe system at a time prior to the lower limit of integration. For example, what was the velocity and height of the helicopterprior to t=10?
49
 
Best answer
2021-10-07

Step 1

Given:

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of \(5.0 m/s^2\). Secret agent Austin Powers jump on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s.

Step 2

(a)

The value of the maximum height that gets reached by the helicopter is determined based on finding the height those results in running of the engine. The additional distance is calculated based on the residual velocity that results in making the engine free fall.

The height being reached by the helicopter on which the engine is running is calculated as below,

\(d=v_it+\frac{1}{2}at^2\)

\(=0+(0.5)(5.0 \ m/s^2)(10.0 s)^2\)

=250 m

After Power shuts off the engine, the helicopter will continue to climb until its velocity is 0 m/s.

The additional distance the helicopter Climbed is,

\(v^2=v_0^2+2ad'\)

\(0=[(5.0 m/s^2)(10.0 s)]^2+2(-9.81 m/s^2)d'\)

\(d'=127.5 m\)

Step 3

The maximum height reached by the helicopter is the sum of the height that it reaches when its engine is running and the additional distance that it climbs due to its residual velocity in free-fall.

That is,

H=d+d'

=250 m+127.5 m

=377.5 m

\(\approx 380 m\)

b) The time that the helicopter takes to crash to the ground after Power shuts off its engine is,

\(d=d_0+v_i t +\frac{1}{2}at^2\)

\(0=250 m +(50 m/s)t+(0.5)(-9.81)t^2\)

\(4.9t^2-50t-250=0\)

Step 4

This is in the form of quadratic equation, using the quadratic formula to solve for t, we get t = 13.87 s

Power steps out of the helicopter immediately after he shuts off the engine. And he deploys a jet pack 7.0 s after leaving the helicopter, with a downward acceleration of \(2.0 m/s^2\). After leaving the helicopter, he is at a height of,

\(d=d_0+v_i t+\frac{1}{2}at^2\)

\(=250 m +(50 \ m/s)(7.0 s)+(0.5)(-9.81)(7.0s)^2\)

d=360 m

Velocity of the Power after leaving the helicopter is,

\(v=v_0+at\)

\(=50 m/s +(-9.81 m/s^2)(7.0 s)\)

=-18.67

Step 5

Velocity of the Power after leaving the helicopter is,

\(v=v_0+at\)

\(=50 m/s +(-9.81 m/s^2)(7.0 s)\)

=-18.67

The amount of time that he uses the jetpack is,

\(t'=13.87-7=6.87 s\)

After Power activates his jetpack, he is at a height of,

\(d=d_0+v_i t+\frac{1}{2}at^2\)

\(=360 m +(-18.67 m/s)(6.87 s)+(0.5)(-2.0 m/s^2)(6.87)\)

=185 m

When the helicopter crashes t0 the ground, Powers is 185 m above the ground.

17

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