#1**+10 **

Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have

h' =-32t + v_{0}

And at the max ht, this is = 0, so we have

-32t + v_{0} = 0 → v_{0} = 32t ...and, substituting this back into the position function at the time of the max ht, we can find the time it takes to reach that point...so we have

1081 = -16t^2 + (32t)t + 921 simplify

1081 = -16t^2 + 32t^2 + 921

160 = 16t^2 divide both sides by 32

10 = t^2

t = √10 sec ..and this is the time it takes to reach the max ht

So, using the first derivative to solve for v_{0} when t= √10 , we have ... v_{0} = 32(√10)ft/s

And our original position function becomes

h = -16t^2 + 32√10t + 921

Here's the graph of the function.....https://www.desmos.com/calculator/42nlumsshk

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As I'm not great at Physics, could someone else (Alan or Melody??) check my answer ???

CPhillJan 5, 2015

Jan 4, 2015