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