Kinematics question

Question

A helicopter, initially at rest on the ground, rises vertically with constant acceleration. When it is at height of 60 m it's upward speed is 5 m/s. When it is at height of 240 m and still rising , an object A is released from the helicopter. Calculate :

(a) the initial velocity of A

(b) the time that A takes to reach the ground.

After A is released the helicopter continues to rise with a different constant acceleration. When it is at a height of 350 m and rising with a speed of 15 m/s, a second object is released. 

(c) Show that B takes 10 seconds to reach the ground.

(d) find the time that elapses between the impacts of A and of B on the ground. 

Take g = 10 m/s^2

Answer 1

Hi Orawan:

 

   Since there are five kinematic quantities (displacement, initial velocity, final velocity, acceleration, and time), and each of the four major kinematic equations involves four of them, most kinematics problems boil down to knowing three of the kinematic quantities to solve for a fourth (the initial velocity appears in every equation, so it must always either be known or be the quantity being solved for).

 

   So the question is, when A is dropped, which among the kinematic quantities do we want to find, and which do we know over the interval in question (the interval during which A drops)?  I usually start with what we want to find, since it is typically spelled out in the question, although in this case, it take a bit of reasoning.

 

   If A is dropped from the helicopter when the helicopter is at a height of 240 m, then its initial velocity must be the same as the helicopter's at that height.  So its initial velocity is the same as the final velocity of the helicopter after the helicopter has risen 240 m.  So the first problem we want to solve is, what is the final velocity of the helicopter over the interval where it rises 240 m?

 

to find:  final velocity (v)

 

What do we know?  We are told the helicopter starts at rest, so we know its initial velocity (v_o) = 0.  We also know its displacement over this interval, (y-y_o) = (+)240 m.

 

know: v_o, (y-y_o)

 

But after this, we are stuck.  We need to know one more kinematic quantity over this 240 m, and none is given directly.  The two remaining kinematic quantities we have not dealt with are time (t) and acceleration (a).  We are given no information about the time, so that is out.

 

However, we are told the acceleration on ascent is constant (it must be, for the kinematic equations to be valid).  And we are given more information about the interval from ground to 60 m in height.  If we can calculate the acceleration over that interval, w know that it will the same as the acceleration over the first 240 m of rise.  So let's try to calculate the acceleration over the first 60 m of rise.

 

In this case, we have:

 

to find:  a

 

In this case, v_o is still the same, because the helicopter still launched from an initial state of rest on the ground.  Ini addition, we are told the displacement ((+)60 m) and the final velocity (5 m/s; positive, because it is upward).

 

know:  v_o, v, (y-y_o)

 

Knowing three kinematic quantities, we can proceed.  We just have to choose the kinematic equation relating displacement, initial velocity, final velocity, and acceleration.  That would be:

 

v² = v_o² + 2a(y-y_o)

 

If you plug in the known values for v, v_o, and (y-y_o), you can solve for a.  I obtained a = (5/24) m/s²

 

Now we can go back to the 240 m interval we started with, and have:

 

to find:  v

know:  v_o, (y-y_o), a

 

We once again want to use the kinematic equation relating those four quantities.  It is the same one as above, only this time we are solving for v.  Note that, when you solve for v, you are taking a square root, and you have to choose (based on your knowledge of the situation) either the positive or negative square root.  In this case, the problem makes a point of saying the helicopter is still rising (which would also be the logical assumption, given the constant acceleration), so we want to take the positive square root to solve for v.

 

When I did this, I calculated v = 10 m/s as the answer to (a).

 

For (b), we now look over the falling interval from 240 m to ground for A.  We ask the same questions:  which kinematic quantity are we trying to find, and which three do we know?

 

to find:  t

know:  v, a, (y-y_o) 

 

v= 10 m/s (calculated in Part (a))

a = -10 m/s² (because once A is released, it is in free fall -- neglecting air resistance, as we usually do in these problems -- and free fall acceleration is g, downward (negative, as I have defined it)

(y-y_o) = -240 m  (A ends up 240 m below where it started when it reaches the ground, so its displacement vector is downward, which, again, I have defined as negative)

 

Now we look at the kinematic equations, and choose the one that connects those four quantities.  It is:

 

(y-y_o) = v_ot + (1/2)at²

 

We can put in the known quantities and solve for t.  For me, this involved using the quadratic formula, and I obtained an answer t = 7.93 s (there is also a negative answer for t, which we can dismiss as "unphysical;"  nothing in the kinematic equations prevents time from running backwards, but that is not normally how we experience time).

 

In Part (c), you know and want to find the same kinematic quantities, just over a different interval (the interval for the helicopter to get to 350 m in height).  They even give you the s final speed there, so you can follow exactly the same process to get the time it takes B to reach the ground.  When I did this, I got the answer they wanted you to show.

 

Then, in Part (d), you need to calculate the difference in the impact times.  We already know how long it takes A and B to reach the ground.  Between A and B reaching the ground, the helicopter also rises from 240 to 350 m.  

 

So from the point A is released:

 

t_A (time for A to hit ground) = 7.93 s (calculated in Part (b))

t_B (time for B to hit ground) = t_rise+10 s (the time for the helicopter to rise from 240 m to 350 m plus B's fall time, given in Part (c))

 

And the difference in those times is:

 

(t_rise+10 s) - (7.93 s)

 

So, if t_rise can be calculated, that solves the problem.  So, let's look at that interval:

 

to find:  t

know: (y-y_o), v_o, v, a

 

(y-y_o) = (+)110 m

v_o = 10 m/s (calculated in Part (a); it has this velocity at 240 m in height)

v = 15 m/s (given in Part (c))

a = (5/24) m/s²  (calculated as a step in Part (a))

 

In this case, we are awash in knowledge! We know all the kinematic quantities, besides t, over this interval.  We can use pretty much any kinematic equation that involves t solve for it.  I choose the one which involves the least algebra:

 

v = v_o+at

 

Using that equation, I calculated t_rise = 48 s.

 

Putting this into the time difference equation above, you can now solve for the time interval between A and B reaching the ground.

 

If you have any other questions, just let me know!