Question

Hello, I have a question about Lecture Clip#2 (Weight and Weightlessness). At around 42:23 (time in the lecture clip), Professor Lewin calculates the maximum height to be 900meters. I also tried to calculate this by setting the y-component of velocity to zero, but I got a different answer. Can someone please explain how to get the 900meters and 30 seconds answer? Thanks in advance!
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-7/
(I've posted the direct link to the video above)

Answer 1

Okay, the professor made it pretty confusing.

You have to remember that at the start of the problem he states the speed in MILES PER HOUR. Mile is not a metric unit so you will have to do some computations. Now, we assume a mile to be 1609 meters, and let's do the conversion:

300 * 1609 = 482700

Now we have 482700m/h, but let's make is m/s to make the calculations easier.

482700 / 3600 = 134

So 300 miles per hour is equal to 134m/s. Now we can handle the problem much easier!

To get the maximum height, we can just ignore the x-velocity and focus on the y-velocity as one dimensional motion. We will use this formula:

d = vf^2 - vi^2 / 2a

Note that final velocity is zero and acceleration is the gravity. Plugging in the values...

d = -(134^2) / 2(-10) = 897.8

897.8 seems close enough to 900 meters right? :P

And the crude calculation of the time proves to be around 14 seconds. I guess the professor made it 15 to look a little better, so no need to sweat about that haha. Hope this helped!

Answer 2

I now understand that I need to convert mph to meters/second. However, I do not get your equation "d=vf^2 - vi^2 / 2a".

What I did was:
To calculate time, I used: v_t= v_0 + at
0=134.11 - 9.8t
t=13.68sec, which i think matches your 14 sec

Then, to calculate the max height of the parabola, i used:
Height = y_0 + (v_y,0)(t) + (1/2)(a)(t^2)
= 0 + (134.11)(13.68) - (1/2)(9.8)(13.68^2)
= 917. 63 meters

Answer 3

For one, I used 10 instead of 9.8 for gravity, since we're just aiming for crude values here. And as to my equation, here's a reference: http://en.wikipedia.org/wiki/Equations_of_motion#Applications

It's the first equation you see. I used it because initially, no time was given but we had initial and final velocity and the acceleration (gravity).

Answer 4

Ok, thanks!