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Mathematics 8 Online
OpenStudy (anonymous):

How long does it take an acorn to hit the ground after dropping from the branch of a tree 12.5 m high? Remember that g=9.8 m/s^2! 1.0 s, 5.0 s, 2.6 s, or 1.6 s **not sure about this!! :(

jimthompson5910 (jim_thompson5910):

what variables do you know

jimthompson5910 (jim_thompson5910):

ie what is given to you

OpenStudy (anonymous):

distance=12.5 m ?

jimthompson5910 (jim_thompson5910):

so d = 12.5

jimthompson5910 (jim_thompson5910):

what else

OpenStudy (anonymous):

initial velocity would be 0?

jimthompson5910 (jim_thompson5910):

yes, vi = 0

OpenStudy (anonymous):

acceleration=9.8 m/s^2 ?

jimthompson5910 (jim_thompson5910):

yes a = 9.8

OpenStudy (anonymous):

not sure if there's anything else though!

jimthompson5910 (jim_thompson5910):

look along the top row of the chart http://www.studyphysics.ca/newnotes/20/unit01_kinematicsdynamics/chp04_acceleration/images/formulas.GIF

OpenStudy (anonymous):

is final velocity known? :/

jimthompson5910 (jim_thompson5910):

what letters or symbols do they have

OpenStudy (anonymous):

we have a, d, vi but missing vf and t?

jimthompson5910 (jim_thompson5910):

what variable do we want to solve for

OpenStudy (anonymous):

t?

jimthompson5910 (jim_thompson5910):

we don't need vf, so look for an X under vf and you'll find your formula

OpenStudy (anonymous):

okie! so d=vi(t)+1/2(at^2) 12.5=0(t)+1/2(9.8t^2) =12.5=4.9*t^2/2 ? not sure if i did that last part right? :/

jimthompson5910 (jim_thompson5910):

you should have 12.5 = 4.9t^2

OpenStudy (anonymous):

okie! ohh okay:) so divide 4.9 right? and you get 2.55102040816=t^2 ?

jimthompson5910 (jim_thompson5910):

then take the square root of both sides

OpenStudy (anonymous):

so the answer will be about t=1.6 s ?

jimthompson5910 (jim_thompson5910):

t >= 0

jimthompson5910 (jim_thompson5910):

t = 1.59719 or yeah, roughly t = 1.6 seconds

OpenStudy (anonymous):

ahh yay! thank you!!!

jimthompson5910 (jim_thompson5910):

np

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