Rocketeer to the rescue: kinematics

[leprofece]

A determined student is dropped from the CN Tower in Toronto, 553 in, and fall librernente. Its initial velocity is zero. Rocketeer arrives 5.00 seconds after and it fall of the tower to save him. Rocketeer is has a velocity down of magnitude vo. In order to avoid injury. Rocketeer must catch at student at a height so that it can stop and arrive at ground at zero speed. the acceleration up to succeed comes from the Rocketeer rocket. which turns right when caught the student; before. This free fall. Not to hurt students, the magnitude of the acceleration of the Rocketeer and student to go down together should be no more than 5 times g. what is the minimum height on the ground which Rocketeer must catch the student?book said the distance fallen in free fall is 5 times the distance from the ground when caught and so the distance is one sixth the toower height or 92.2 m
But it says so without no calculation
I don't have any idea of where the 1/6 comes from
Can anyone help me??'
thanks

 

[MarkFL]

Re: physics 2

I moved this thread as it is algebra based kinematics and gave it a descriptive title.

Let's begin with the kinematic relationship:

\(\displaystyle a=\frac{v_i^2-v_f^2}{2\left(y_i-y_f \right)}\)

and apply it to the period of time the student and the Rocketeer are slowing down. We are told $a=5g,\,v_f=0,\,y_f=0$ and so what do we have?

 

[zzephod]

A determined student is dropped from the CN Tower in Toronto, 553 in, and fall librernente. Its initial velocity is zero. Rocketeer arrives 5.00 seconds after and it fall of the tower to save him. Rocketeer is has a velocity down of magnitude vo. In or...

Asked before http://mathhelpboards.com/advanced-applied-mathematics-16/checking-work-2-71-university-physics-warning-there-lot-5350.html, here, here and here and probably many other places.

.

 

[leprofece]

Re: physics 2

I moved this thread as it is algebra based kinematics and gave it a descriptive title.

Let's begin with the kinematic relationship:

\(\displaystyle a=\frac{v_i^2-v_f^2}{2\left(y_i-y_f \right)}\)

and apply it to the period of time the student and the Rocketeer are slowing down. We are told $a=5g,\,v_f=0,\,y_f=0$ and so what do we have?

5g = 0-vf²/2(0-553)
5(10)(1106)= vf²
sqrt(55300) = vf

vf = 235,15 m/s

if plug it into y = vo²/(2.(5g)) I don't get the right one so how?

 

[MarkFL]

Let's begin with the kinematic relationship:

\(\displaystyle a=\frac{v_i^2-v_f^2}{2\left(y_i-y_f \right)}\)

and apply it to the period of time the student and the Rocketeer are slowing down. We are told $a=5g,\,v_f=0,\,y_f=0$ and so we have:

\(\displaystyle 5g=\frac{v_i^2}{2y_i}\)

Now, we want to solve for $y_i$, the height at which the Rocketeer catches the student.

\(\displaystyle 10y_i=\frac{v_i^2}{g}\)

Now, we also know that \(\displaystyle v_i=gt\) and so we have:

\(\displaystyle 10y_i=gt^2\)

Let $h$ be the height of the tower, and we also know that \(\displaystyle y_i=h-\frac{gt^2}{2}\,\therefore\,gt^2=2\left(h-y_i\right)\) which gives us:

\(\displaystyle 10y_i=2\left(h-y_i\right)\)

\(\displaystyle 5y_i=h-y_i\)

\(\displaystyle y_i=\frac{h}{6}\)