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?
not doing your homework
Better
Can't do it can you?
The problem is really fucking easy. If you think about it for more than a second.
I already solved it my man
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
wrong
>using the imperial system for physics problems
Amerifats are truly disgusting
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.
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
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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
Post the book solution so we can explain why they did it that way.
right
Sorry I see now, while the ball falls the other bojects velocity would increase
nevermind I have no idea what's going on someone please explain
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
Oh i get it
nvm, I just had to use my pea brain a little
Mods do your goddamn jobs
At break neck speeds
>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?
Post book solution, I want to see this done with integrals
it's here
>integrals
user this is literally pre high school physics
taking a derivative and turning into a formula is integration. it's an integral. the question itself is just an arbitrary example.
HAHA NICE, user, SHOEHORNING POLITICS INTO ANYTHING MAKES IT IMMEDIATELY BETTER XXXXDDDDDDDDDDDDD
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Fucking bitch, post the book solution. We all helped and you fucking can't do that?
Fuck you!!!