# Physics dealing with distance

The distance $s$ in feet that a thrown object will fall in $t$ seconds is given by

$$s=16t^2 + vt\;,$$

where $v$ is the initial downward velocity in feet/second. Find the initial velocity of an object that falls $80$ feet in $2$ seconds.

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What have you done? Where are you stuck? –  Pragabhava Nov 13 '12 at 1:14

Hint: you have three unknowns in the equation and are provided values for two of them.

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So you have

• $s =$ distance
• $v =$ initial velocity
• $t =$ time.

You have these related via: $$s = 16t^2 + tv.$$

Now if the distance $d= 80$ (feet) and the time is $t = 2$ (sec.), you know two of the three variables, so you can solve for the third. I.e., start with $$80 = 16(2)^2 + 2v.$$ This is a (linear) equation with just one unknown that you can solve for.

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