An object is thrown down from the top of a 480 foot building with an initial velocity of 64 ft/sec.
Find position and velocity after 2 seconds
A. What is its position after 2 seconds?
B. What is its velocity after 2 seconds?
C. How long until it hits the ground?
The formula for the position function when measured in feet is:
`s=-16t^2+v_0t+s_0` where `t` is time measured in seconds, `v_0` is the initial velocity measured in `(ft)/s` , and `s_0` is the initial height measured in feet.
The velocity function is the derivative of the position function. (It measures the rate of change of position with respect to time)
Thus `v=-32t+v_0` .
(1) The position at time `t=2` is
(2) The velocity at time `t=2` is
(3) To find when it hits the ground, set `s=0` , thus:
`t=(-4 +- sqrt(16-4(1)(-30)))/(2)~~3.83`
Thus it hits the ground after approximately 3.83 seconds.