Suppose a squirrel drops an acorn from a branch at a height of 60ft.?

What would express the height of the acorn above the ground as a function of time?





Suppose a squirrel drops an acorn from a branch at a height of 60ft.?
t = time


d = distance (height)


a = acceleration (32 ft/s虏)





d = 1/2 * a * t虏


60 f = 1/2 * 32 f/s虏 * t虏


t虏 = 60 f * 2 / 32 f/s虏


t虏 = 3.75 s虏


t = 1.93649167 s


t = ~2 s to 1 significant digit.





To express height as a function of time:


h = 1/2 at虏


t虏 = 2h/a


t = 鈭?h/32


t(h) = 鈭歨 / 4





Edit: You can graph ANYTHING, not just x and y. If you insist on x or y, then f(x) = 鈭歺 / 4 means the same thing, but t(h) = 鈭歨 / 4 IS expressing height as a function of time.Suppose a squirrel drops an acorn from a branch at a height of 60ft.?
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
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 %26gt;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!
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h = 60-(32.2*t^2)
y=60-5t^2