The problem is as follows:

In a certain shopping mall which is many stories high there is a glass elevator in the middle plaza. One shopper ridding the elevator notices a kid drops a spheric toy from the top of the building where is located the toy store. The shopper riding the elevator labeled $A_{1}$ is descending towards the ground with a velocity of $\vec{v}=-5\hat{j}\,\frac{m}{s}$. Find the speed (in meters per second) and the acceleration in $\frac{m}{s^{2}}$ which will be seen by the shopper in the glass elevator in the instant $t=3\,s$. You may use $g=10\,\frac{m}{s^{2}}$

The given alternatives on my book are as follows:

$\begin{array}{ll} 1.&-35\hat{i}-10\hat{j}\frac{m}{s}\\ 2.&-25\hat{i}-10\hat{j}\frac{m}{s}\\ 3.&-30\hat{i}-10\hat{j}\frac{m}{s}\\ 4.&-25\hat{i}+10\hat{j}\frac{m}{s}\\ 5.&-40\hat{i}-10\hat{j}\frac{m}{s}\\ \end{array}$

For this problem I'm totally lost at how should I understand or calculate the speed as seen from the observer. My first guess is that it might be the sum of the two speeds?. In other words that the speed of the shopper inside the glass elevator is the sum of the sphere as seen by him and the real speed. Or could it be the opposite?.

I'm still confused at this part.

The only thing which I could come up with was to write the position equation for the sphere as shown below:


Although $v_{oy}=0$, and $t=3\,s$ there is no given information about how high is the building.

Then I turned my attention to the speed at $t=3\,s$ this would mean:



That would be the real speed of the sphere at that instant. My intuition tells me that the observer will see the ball going faster? and how about the acceleration?

Then and more importantly how can I find the acceleration and the velocity as seen by the observer riding in the elevator?. Can somebody help me here?.

  • $\begingroup$ setting $v_{oy}$ = (initial velocity of sphere) = -(velocity of the elevator) should do $\endgroup$
    – AgentS
    Nov 5 '19 at 4:15
  • $\begingroup$ @pooja Would you perhaps develop your answer a little bit more so I can follow the steps correctly?. How can I justify what you're implying?. Is the descending acceleration additive?. $\endgroup$ Nov 5 '19 at 4:24
  • $\begingroup$ Elevator is going at constant speed right? $\endgroup$
    – AgentS
    Nov 5 '19 at 4:24
  • $\begingroup$ @pooja It doesn't say. But I assume that might be the case. It doesn't make sense the elevator is moving with an acceleration down to the ground. $\endgroup$ Nov 5 '19 at 4:26
  • $\begingroup$ Before the sphere dropped, in observer's frame, the sphere appears to move up with 5m/s. $\endgroup$
    – AgentS
    Nov 5 '19 at 4:27

That would be the real speed of the sphere at that instant. My intuition tells me that the observer will see the ball going faster? and how about the acceleration?

No. Since both the sphere and the observer are going down, the observer will see the ball going slower. As an example, suppose you're chasing me and our speeds are same, then you wouldn't see me moving at all.

Set $v_{oy}$ = (initial velocity of sphere) = -(velocity of the elevator) and work the kinematics equations as usual.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Nov 5 '19 at 9:47
  • $\begingroup$ @pooja Mind checking? I did updated the chat with a clarification regarding my comment and answer to your latest comment as well. $\endgroup$ Nov 7 '19 at 8:16

If the elevator is moving at a constant velocity, the acceleration due to gravity is the same inside the elevator as outside.

$a = -10\frac {m}{s^2}$

Velocities will be different due to the different frames of refference.

For the person in the elevator the relative velocity is.

$v = (-10t + 5) \frac{m}{s}$ at time $t=3, v(t) = -25\frac {m}{s}$

  • $\begingroup$ Okay where does the $-10t$ comes from?. $\endgroup$ Nov 5 '19 at 4:34
  • $\begingroup$ Is it because of the acceleration due gravity?. $\endgroup$ Nov 5 '19 at 4:35
  • $\begingroup$ Gravity provides acceleration to falling bodies. $\endgroup$
    – Doug M
    Nov 5 '19 at 4:36
  • $\begingroup$ Yes I know that. What I believe you might put is the equation in this order $v= +v_{o} - gt=+5-10t$ that's much easier for me to spot what you were intending to refer. $\endgroup$ Nov 5 '19 at 4:38
  • $\begingroup$ Addition is commutative. $\endgroup$
    – Doug M
    Nov 5 '19 at 4:42

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