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 November 4th, 2019, 05:13 PM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus How do I find the velocity of a sphere falling when an observer riding an elevator? 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 riding 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 by this part. The only thing which I could come up with was to write the position equation for the sphere as shown below: $y(t)=y_{0}+v_{oy}t-\frac{1}{2}gt^2$ Although $v_{oy}=0$, and $t=3\,s$, there is no given information about how high the building is. Then I turned my attention to the speed at $t=3\,s$ this would mean: $v_{f}=v_{o}-gt$ $v_{f}=0-10(3)=-30\,\frac{m}{s}$ 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? Last edited by skipjack; November 4th, 2019 at 10:11 PM. November 4th, 2019, 05:56 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 None of the answer choices given (all velocities) make any sense ... how can two objects, an elevator moving downward and a falling sphere moving downward, have components in the horizontal direction (the $\hat{i}$ in each choice)? Methinks the answer choices were typed incorrectly. $v_e = -5\hat{j} \, m/s$ $v_s(3) = 0 -30 \hat{j} \, m/s$ $v_s - v_e = -30\hat{j} - (-5\hat{j}) = -25 \hat{j} \, m/s$ as far as acceleration is concerned, the elevator is moving at a constant speed ... $a_e=0$ $a_s - a_e = -10 \hat{j} - 0 = -10 \hat{j} \, m/s^2$ Thanks from topsquark and DarnItJimImAnEngineer Tags elevator, falling, find, observer, riding, sphere, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chemist116 Physics 5 November 7th, 2019 01:23 AM pianist Algebra 1 April 14th, 2016 12:47 AM Cartesio Advanced Statistics 2 September 11th, 2014 09:04 AM Melody2 Algebra 2 October 23rd, 2013 05:39 PM joop Linear Algebra 0 July 6th, 2009 03:12 AM

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