Ask your own question, for FREE!
Mathematics 51 Online
OpenStudy (anonymous):

What formula is used for: Object dropped off a 210 ft building. Initial v=-24. What is the velocity at 130 ft? s(t) = -16t^2-24t+210. v(t) = -32-24. Already got velocity after 3 sec but not sure how to adjust formula for v at 130 ft.

OpenStudy (anonymous):

Tried making s(t)=0 to find t then install value for t into v(t)=-32t-24 but polynomial was not factorable. Quad. equation gave a ridiculous answer.

OpenStudy (anonymous):

-16t^2 - 24t+210=130 Solve for t when you obtain t use the t as substitute for v=32-24

OpenStudy (anonymous):

Your velocity formula is wrong btw it should be v(t)=-32t-24

OpenStudy (anonymous):

any question?

OpenStudy (anonymous):

can't solve for t b/c haven't been able to factor (-16t^2-24+80)=0. quad equation gives t= about 2.9 which doesn't make any sense b/c it isn't an option. And, check back. My v formula is v(t)=-32-24

OpenStudy (anonymous):

it is wrong, it should be -32t-24

OpenStudy (anonymous):

assuming that t=2.9 is right I would plug into my V(t) -32(2.9)-24=

OpenStudy (anonymous):

Ok, must not have hit the t key hard enough but I have been using the t. I get -116.8 but the answer is -94.32.

OpenStudy (anonymous):

I did the quad equation twice. But id that is the correct formula then I guess that is all I need to know. I will do it again.

OpenStudy (anonymous):

This iswhat I got using quad equation. Was trying to remember what I got for t last night (not 2.9 it is 2.36). Anyway, either way the answer is not right and it is a ridiculous answer so if I did it right then I don't get how -16t^2 - 24t +80 is even factorable b/c the right answer is -94.32. -16t^2 -24t+80=0 -(-24) +- (Sqrt [(-24)^2]-4(-16)(80)/2(-16) 24 +- (sqrt 576+5120)/-32 24 +- (sqrt 5696)/-32 t= 24-2.36 or 24 + 2.36 -32(21.64)-24= -716.48 or -32(26.36)-24= -867.52

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!