Superman Problem: Catching a Falling Student

[Edwardo_Elric]

Homework Statement

Determined to test the law of gravity for himself,a student walks off a skyscraper 200 m high, stopwatch in hand and starts his free fall (zero initial velocity. Five seconds later, superman arrives at the scene and dives off the roof to save the student.
a.) What must be the magnitude of Superman's initial velocity so that he catches the student just before the ground is reached? (Assume superman's acceleration is that of any freely falling body. Superman leaves the roof with an initial speed Vo that he produces by pushing downward from the edge of the roof with his legs of steel).

b.) If the height of a skyscraper is less than some minimum value, even superman can't reach the student before he hits the ground. What is the minimum height?.

Homework Equations

y = Vot - 1/2(gt^2)

The Attempt at a Solution

t = student
t+5 = when superman arrives
Stuck with letter a
a.)
First i equate the two y positions:
y_{superman} = y_{student}

y_{student} = Vot - 1/2(gt^2)
y_{student} = - 1/2(gt^2)

y_{superman} = Vo(t + 5s) - 1/2(g(t+5s)^2)
y_{superman} = Vo(t+5s) - 1/2(9.8)(t^2+10t + 25)
= Vo(t+5s) - 1/2(9.8)t^2 - 5(9.8)t - 122.5t

Equate
y_{superman} = y_{student}
Vo(t+5s) - 1/2(9.8)t^2 - 5(9.8)t - 122.5t = - 1/2(9.8)(t^2)

cancel -1/2(9.8)t^2

Vo(t+5s) = 171.5t

Vo = 171.5t /(t+5s)

 

[HallsofIvy]

Homework Statement

Determined to test the law of gravity for himself,a student walks off a skyscraper 200 m high, stopwatch in hand and starts his free fall (zero initial velocity. Five seconds later, superman arrives at the scene and dives off the roof to save the student.
a.) What must be the magnitude of Superman's initial velocity so that he catches the student just before the ground is reached? (Assume superman's acceleration is that of any freely falling body. Superman leaves the roof with an initial speed Vo that he produces by pushing downward from the edge of the roof with his legs of steel).

b.) If the height of a skyscraper is less than some minimum value, even superman can't reach the student before he hits the ground. What is the minimum height?.

Homework Equations

y = Vot - 1/2(gt^2)

The Attempt at a Solution

t = student
t+5 = when superman arrives
Stuck with letter a
a.)
First i equate the two y positions:
y_{superman} = y_{student}

y_{student} = Vot - 1/2(gt^2)
y_{student} = - 1/2(gt^2)

y_{superman} = Vo(t + 5s) - 1/2(g(t+5s)^2)
y_{superman} = Vo(t+5s) - 1/2(9.8)(t^2+10t + 25)
= Vo(t+5s) - 1/2(9.8)t^2 - 5(9.8)t - 122.5t

Equate
y_{superman} = y_{student}
Vo(t+5s) - 1/2(9.8)t^2 - 5(9.8)t - 122.5t = - 1/2(9.8)(t^2)

cancel -1/2(9.8)t^2

Vo(t+5s) = 171.5t

Vo = 171.5t /(t+5s)

You haven't used the fact that the skyscraper is 200 m high and superman catches him "just before the ground is reached".

 

[PhanthomJay]

You are making the problem more complex than it need be. Calculate how long, T, it will take the student to just hit the ground. Then superman, leaving 5 seconds later, must now dive the 200m in (T -5) seconds. Your minus sign in the kinematics equation is also not correct. I am assuming that superman has an initial speed down, same as the displacemnt direction, same as the acceleration direction, all values should be plus.

 

[Edwardo_Elric]

so that means he is luckily catching him before striking the ground
a.)Calculating for time of student to reach the ground:
y = 1/2(gt^2)
sqrt(((2)(200m))/2) = t
t = 6.3s
Superman's initial speed:
(t - 5s);

Vo(t-5) - 1/2(g(t-5s)^2) = 200
Vo(6.3-5) = 200 + 8.281
Vo = 208.281/(6.3-5)
Vo = 160 m/s => superman speed

b.) minimum height
i used the derivative of y
Vo(t-5) - 1/2(g)(t-5s)^2 = y
Vo - g(t-5) = dy/dt
160 - 9.8(1.3s)= dy/dt
147.26m/s = dy/dt

substitute

y = Vot - 1/2(g(t-5.0s)^2)
= 147.26 - 1/2(9.8(1.3s)^2)
= 138.979m
minimum height

 

[PhanthomJay]

I had a feeling that the plus or minus signs would be troubling in this problem. You are messing up your plus and minus signs in your kinematics equation, in part because the problem was not clearly worded, and in part because you are not watching your directions of the velocity, acceleration, and displacement vectors. The problem perhaps should have been better stated by noting that superman's initial velocity is DOWN (which he achieves by pushing UP on the roof edge while diving head first, by the way, but that's another source of confusion). So in your kinematic equation for superman, if you choose down as the negative direction, which you have done by using 'g' as a negative value, you should then also put a minus sign in front of the 200. Or else, I find it easier to consider down as the positive direction, and make all your values positive. So it's either
-200 = V_ot - 1/2gt^2 or
200 = V_ot + 1/2gt^2
where t = (6.3 - 5), which you have calculated corectly. Solving either of these equations will not equate to your result for V_o.
For part b, it's easy for anyone to lose me with the calculus stuff. I'd opt to note that while superman may travel faster than a speeding bullet, and maybe perhaps even faster than light, he can't travel at an infinite speed. So if the student hits the ground in exactly 5 seconds, it's too late for anyone leaving 5 seconds later to save him. At what building height would the student hit the ground in 5 seconds?

 

[t_nhung]

part B.
simply plug in t=5 into the height equation for student to find min height, which is:
y = 0.5 * g * t
y = 0.5 * 9.8 * 5
y = 123 (m)

(t = 5 because Superman only arrives 5 seconds later; that means if the student hit the ground in 5 seconds, Superman won't be able to save him.)