# 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 secondsA. What is its position after 2 seconds?B. What is its...

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?

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### 1 Answer

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` .

So:

(1) **The position at time `t=2` is**

`s=-16(2)^2-64(2)+480=280` feet.

(2) **The velocity at time `t=2` is**

`v=-32(2)-64=-128 (ft)/s`

(3) To find when it hits the ground, set `s=0` , thus:

`0=-16t^2-64t+480`

`0=-16(t^2+4t-30)`

`t=(-4 +- sqrt(16-4(1)(-30)))/(2)~~3.83`

**Thus it hits the ground after approximately 3.83 seconds.**