Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Mar 5th 2011, 08:17 AM #1 Junior Member   Join Date: Nov 2010 Location: Plymouth, MA Posts: 22 Ball drops and ball tthrow at same time.. Ball dropped from the top of a building at the same time another ball is thrown upward from the ground with Vo of 40 m/s....velocities of both balls are the same when they reach the center of a 4m window. what is the height to the top of the window which the ball has dropped. The window is 20 meters from he ground. got 20.5 meters for the height from the top of window to top of roof where ball was dropped. #2) Ball is dropped , at rest, from top of a building...takes twice as long for the ball to drop off the last 500m as it does for the ball to drop the first 'h' meters | So height of the building is equal to 500+h...what is h? I think I needed to figure out the times (T1 and T2)...Im kind of stumped here When you gfuys answer could you leave a somewhat detailed summary...   Mar 5th 2011, 02:32 PM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,667
 Originally Posted by LittleWing Ball dropped from the top of a building at the same time another ball is thrown upward from the ground with Vo of 40 m/s....velocities of both balls are the same when they reach the center of a 4m window. what is the height to the top of the window which the ball has dropped. The window is 20 meters from he ground. got 20.5 meters for the height from the top of window to top of roof where ball was dropped. #2) Ball is dropped , at rest, from top of a building...takes twice as long for the ball to drop off the last 500m as it does for the ball to drop the first 'h' meters | So height of the building is equal to 500+h...what is h? I think I needed to figure out the times (T1 and T2)...Im kind of stumped here When you gfuys answer could you leave a somewhat detailed summary...
(snorts) I think the real problem for this question doesn't involve Physics. It's in trying to find out what the heck the problem is about....

Okay, if I understand this correctly you have a building of height h. The ball at the top is going to drop a distance d to get to the top of the window, then another 2 m at which point the ball at the top and the ball from the bottom have the same speed. The window is 20 m high, so this point is 22 m high.

Thus we know the initial speed of the ball that's being thrown upward (40 m/s) and we know it has to get to 22 m, at which the speed of this ball is the same as the speed as the one that gets dropped. Let's call this speed v. Using
v^2 = v0^2 + 2a(y - y0)
I get that the speed v is 34.19 m/s.

Using that as the final speed for the ball that got dropped (and using the same equation for this as the ball that was thrown upward) I get d = 61.63 m as the distance from the top of the building to the top of the window.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.   Mar 7th 2011, 09:08 AM #3 Junior Member   Join Date: Nov 2010 Location: Plymouth, MA Posts: 22 Yes I know my physics professor is very vague with his questions..thank you for answering and decyphering my cryptic question. Here is another attempt at the second question.... Ball is dropped from a building and it takes twice as long for the ball to drop off the last 500m as it does for the ball to drop the first 'h' meters. Determine H..... T(2)= T(1) x 2   Mar 7th 2011, 11:03 AM   #4

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,667
 Originally Posted by LittleWing Yes I know my physics professor is very vague with his questions..thank you for answering and decyphering my cryptic question. Here is another attempt at the second question.... Ball is dropped from a building and it takes twice as long for the ball to drop off the last 500m as it does for the ball to drop the first 'h' meters. Determine H..... T(2)= T(1) x 2
I would do this in a few steps (for the sake of clarity.)

Let the height of the building be H. When the ball is at 500 m, let this be at time T. Then we know that
y = y0 + v0*t + (1/2)at^2 gives:

500 = H - (1/2)*gT^2

We need another equation in H and T to solve this. So let's take the second part. The initial time is T and the final time is T + 2T = 3T. The ball starts at y0 = 500 m. So we have
y = y0 + v0*(t - t0) + (1/2)a(t - t0)^2

(This equation is the same as y = y0 + v0*t + (1/2)at^2 with the exception of a "displaced" time variable.)

So
0 = 500 + v0*(2T) - (1/2)g*(2T)^2

Notice that the v0 in this equation is equal to the final v in the first part. So we need v from the first part:
v = v0 + at

v = -gT

Putting this v into the equation for the second part of the fall we get as v0:
0 = 500 + (-gT)*(2T) - (1/2)g*(2T)^2

This is an equation for T. From here we can get H:
500 = H - (1/2)gT^2

and then h:
H = h + 500

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.  Tags ball, drops, time, tthrow Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post johann1966 Kinematics and Dynamics 2 Feb 17th 2016 12:33 PM teddybear9 Kinematics and Dynamics 2 Nov 15th 2011 12:31 PM Pip Threlfall Kinematics and Dynamics 10 May 29th 2009 12:46 PM ariol Kinematics and Dynamics 1 Jan 9th 2009 06:04 AM werehk Kinematics and Dynamics 1 Jun 18th 2008 04:14 AM 