Complicated Kinematic problem dealing with falling

[GZM]

Homework Statement

Problem states : A student walks off the top of the CN Tower in Toronto, which has height h and falls freely. His initial velocity is zero. The Rocketeer arrives at the scene a time of t later and dives off the top of the tower to save the student. The Rocketeer leaves the roof with an initial downward speed v0. In order both to catch the student and to prevent injury to him, the Rocketeer should catch the student at a sufficiently great height above ground so that the Rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the Rocketeer's jet pack, which he turns on just as he catches the student; before then the Rocketeer is in free fall. To prevent discomfort to the student, the magnitude of the acceleration of the Rocketeer and the student as they move downward together should be no more than five times g.

It asks for : What is the minimum height above the ground at which the Rocketeer should catch the student?
and :
What must the Rocketeer's initial downward speed be so that he catches the student at the minimum height found in part (a)?
(Take the free fall acceleration to be g.)

Homework Equations

Kinematic equations :
d=vi*t+(1/2)*a*t^2
vf^2 = vi ^2 + 2*a*d
vf = vi + at

The Attempt at a Solution

My attempt of the solution was this diagram, but I don't know what I can equate other than d to each other to solve, but that won't give me anything for the distance that the Rocketeer catches the student, but all I know is the d is equivalent for both of these guys

 

[andrewkirk]

Solve part (a) first. You will get two equations with two unknowns. For instance the two unknowns could be the height $h_c$ at which the student's 5g deceleration starts and $t_c$ could be the time at which that happens. Then you can write two equations using those unknowns and the known variables, one equation for the free fall and one equation for the 5g deceleration phase. You can solve the system to find $h_c$ and $t_c$.

Then you just need to find the initial downwards velocity of the rocketeer in order for him to arrive at height $h_c$ at time $t_c$ under free-fall acceleration, taking account that he started later than the student by given delay $\delta t$.

 

[Herman Trivilino]

Let's start with your equation

$d=\frac{1}{2}at_1^2.$

$\frac{1}{2}at_1^2$ is the distance $d$ that the student falls during the time $t_1$ before the Rocketeer catches him. Since the student is in free fall you should be able to tell us the value of $a$ in this expression.

So, that's a start!

Next, let's look at your other equation.

$d=\frac{v_2^2-v_i^2}{2a}.$

This is the same distance $d$, the distance the Rocketeer free falls to catch the student, but what you're calling $v_i$ is actually given in the statement of the problem to be $v_o$. Again, you should be able to tell us the value of $a$.

Make those three changes.

Further thoughts ...

Note that $t_1-t$ is the time spent by the Rocketeer catching up to the student, so you have to work that fact into your equations somehow.

And that $h-d$ is the distance the student and Rocketeer travel together (with the jet pack on), so you have to work that into your equations, too.

 

[GZM]

Solve part (a) first. You will get two equations with two unknowns. For instance the two unknowns could be the height $h_c$ at which the student's 5g deceleration starts and $t_c$ could be the time at which that happens. Then you can write two equations using those unknowns and the known variables, one equation for the free fall and one equation for the 5g deceleration phase. You can solve the system to find $h_c$ and $t_c$.

Then you just need to find the initial downwards velocity of the rocketeer in order for him to arrive at height $h_c$ at time $t_c$ under free-fall acceleration, taking account that he started later than the student by given delay $\delta t$.

Would the answer be left in variables or would there be an exact numerical answer? and when you say I'll get 2 equations with 2 unknowns you mean each or 2 unknowns overall, because I get 2 unknowns each and when I equate them to each other i get 2 unknowns in a single expression

 

[andrewkirk]

Would the answer be left in variables or would there be an exact numerical answer?

If you are given the initial height (ie height of CN tower) and use say 9.8 for g, there will be a numeric answer.

For the first part you should get two equations with two unknowns - in total, not each. My unknowns are $h_c$ and $t_c$. It sounds like you have four unknowns. Post your equations so that people can look at them, and help will no doubt be provided.