A body is dropped on Earth. At the instant when it has fallen 100 feet another second body is to be projected downward with an initial velocity which will enable it to catch up with the first body 10 seconds after the second body is released.
What is the requisite initial velocity of the second body?
Aiden White
not doing your homework
Logan Bailey
Better
Levi Nguyen
Can't do it can you?
Leo Butler
The problem is really fucking easy. If you think about it for more than a second.
Caleb James
I already solved it my man
Anthony Miller
g= 32.2feet/s ^2 d=(gt ^2)/2 t=root(2*100feet/32.2feet/s ^2) t1=2.49 s t2=10s+2.49s= 12.49 d2=(gt^2)/2 = (32.2feet/s ^2*(12.49)^2)/2= 2511.60 feet Assuming there is nothing underneath and they sart at the same point V=d/t V=251.16 feet/sec
Lucas Gutierrez
wrong
Evan Ward
>using the imperial system for physics problems Amerifats are truly disgusting
Juan Gutierrez
solve for how long it takes to fall 100 feet. Then add 10 seconds to see how far the thing falls before the 2nd thing catches up. Then solve for starting velocity of 2nd thing given the previously calculated time and distance.
James Baker
Or solve for velocity of the object at 100 feet and add that to 10ft/sec needed to cover the distance between them. 90 ish feet per second
Leo Cox
...
Chase Rivera
That's how I did it, in my head. But, the book shows some other convoluted integrated way that I don't understand because the solution manual is very brief.
It does with only 2 integrals
Hudson Gutierrez
Post the book solution so we can explain why they did it that way.
Nicholas Rivera
right
Aiden Parker
Sorry I see now, while the ball falls the other bojects velocity would increase
Julian Rivera
nevermind I have no idea what's going on someone please explain
Logan King
Is this supposed to be in a "vacuum"? Is drag a factor? body orientation? what's the initial position and velocity of body 1 relative to earth? where does body 2 appear relative body 1 and when? Then tell your TA he is a retard and should pick a different major due to their obvious diminished mental capacity. If your professor wrote this or picked the book it is in, time to transfer schools. Problem solved
Jackson Williams
Oh i get it
Evan Richardson
nvm, I just had to use my pea brain a little
Justin Hughes
Mods do your goddamn jobs
Connor Wilson
At break neck speeds
Nicholas Sullivan
>2511.60 d2=(gt^2)/2 = (32.2feet/s ^2*(12.49)^2)/2= 2511.60 feet which is the distance the first object has fallen that the second object will have to catch up to in 10 seconds d=vi*t+a*t^2/2 vi=(2511.60 feet-(32.2feet/s*10s^2)/2)/10s vi=90.16 feet/sec Am I right now?
James Hill
Post book solution, I want to see this done with integrals
Bentley Cooper
it's here
Samuel Murphy
>integrals user this is literally pre high school physics
Carter Phillips
taking a derivative and turning into a formula is integration. it's an integral. the question itself is just an arbitrary example.
Daniel Sullivan
HAHA NICE, user, SHOEHORNING POLITICS INTO ANYTHING MAKES IT IMMEDIATELY BETTER XXXXDDDDDDDDDDDDD
Isaiah Wilson
...
Landon Murphy
Fucking bitch, post the book solution. We all helped and you fucking can't do that?