Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have
h' =-32t + v0
And at the max ht, this is = 0, so we have
-32t + v0 = 0 → v0 = 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 v0 when t= √10 , we have ... v0 = 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 ???