Suppose a squirrel drops an acorn from a branch at a height of 60ft?
October 10th, 2009
Posted by: admin
What would express the height of the acorn above the ground as a function of time?
the expression has to be graphed so im pretty sure it is an f(x)=….. or y=……
By: melon.flavor
Tags: Acorn, Expression, Squirrel
October 13th, 2009 at 10:59
Hi ,
I am ready to help you. but this questions needs the detailed explanation. Please contact me at. I will provide you the fully explained solution for this question and always ready to help you when ever you require such help in future as well.
Waiting for your reply.
Austin
October 15th, 2009 at 21:27
t = time
d = distance (height)
a = acceleration (32 ft/s²)
d = 1/2 * a * t²
60 f = 1/2 * 32 f/s² * t²
t² = 60 f * 2 / 32 f/s²
t² = 3.75 s²
t = 1.93649167 s
t = ~2 s to 1 significant digit.
To express height as a function of time:
h = 1/2 at²
t² = 2h/a
t = √2h/32
t(h) = √h / 4
Edit: You can graph ANYTHING, not just x and y. If you insist on x or y, then f(x) = √x / 4 means the same thing, but t(h) = √h / 4 IS expressing height as a function of time.
October 16th, 2009 at 06:31
Distance is the integral of the velocity function
Distance = velocity x time
d=vt
if we integrate vt we get
d=(at^2)/2
we know that the gravity of earth is 32.2 m/s^2 so we have
60=(32.2 t^2)/2
you can move the 60 to the other side and get
f(x) = (32.2 t^2)/2 - 60
it will take 1.36 seconds to hit the ground.
v = at so the velocity will be
32.2ft/s^2(1.36s) = 33.56 ft/s
October 17th, 2009 at 00:42
h = 60-(32.2*t^2)
October 18th, 2009 at 14:53
you can figure the distance fallen by
1/2 at^2
so the height at t seconds is 60 - distance fallen
60 - 32t^2/2
=60-16t^2
but remember to ignore negative results,or dont use
time >1.9364916731037084425896326998912
1/2 at^2=60
t=1.9364916731037084425896326998912
and dont look up,silly! That acorn will hit you at 31 ft/sec! Ouch!