Question

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0m/s2. How far is Powers above the ground when the helicopter crashes into the ground?

Answer 1

Answer:

a) 378 m.

b) 184 m.

Explanation:

To obtain the maximun heigth we need to use the formula of accelerated motion:

$Y=Yo+Vo*t+\frac{1}{2}*a*t^2[\tex][tex]Y=0+0*10+\frac{1}{2}*5.0*(10)^2\\Y=250m$

and the velocity of the helicopter when the engine was turned off is given by:

$Vf^2=Vo^2+2*a*dy\\Vf^2=0+2*(5m/s^2)*(250)\\Vf=50m/s$

so it travels a few meter up before going down:

$Vf^2=Vo^2+2*a*dy\\dy=\frac{(0)-(50m/s)^2}{2*(-9.8)}\\dy=128m$

So the maximun height is 378m.

The time the helicopter takes to crash to the ground is given by:

$0=250+50*t+\frac{1}{2}*(-9.80)*(t)^2$

solving the quadratic function: t=13.9s and t=-3.67

we have to take the positive value.

Power is going down with the same acceleration of the helicopter so the distance above the ground after 7 seconds is given by:

$0=250+50*7+\frac{1}{2}*(-9.8)*(7)^2\\Y=360m$

and the velocity at that moment is given by:

$Vf=Vo+a*t\\Vf=50+(-9.8)*7\\Vf=-18.6m/s$

we took the acceleration of the gravity and velocity as negative because the movement is taking place in the negative direction.

the helicopter took 13.9 seconds to crash the ground, so after 6.9 seconds power is at:

$Y=360-(18.6)*6.9+\frac{1}{2}*(-2)*(6.9)^2\\Y=184m$

184 meters above the ground.