ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: fredrick on August 09, 2008, 10:29:12 pm
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[IMG]http://img216.imageshack.us/img216/1297/ugyv1.th.jpg[/img]
:idiot2:
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The speed in which it goes off the cliff in the horizontal direction is irrelevant.
In the vertical direction:
u = 0m/s
a = -9.8m/s^2
t = 3s
height of cliff = ?
Sub numbers into one of the straight line motion equations as accel is constant, namely x = ut + .5at^2
= 0 + .5*-9.8*3^2
The answer is negative because it is indicating the object has moved downwards that distance, so the magnitude of it is cliff height