s= [b]*[d]+64t-16t^2
what is the greatest height above the ground the rock will go
ds/dt = 64 - 32t , assuming b and d are constants
= 0 for a max of s
32t = 64
t = 2
sub back into s
s = bd + 128 - 64
= bd + 64
To find the greatest height above the ground that the rock will go, we need to determine the maximum value of the function s.
Given that s = [b]*[d]+64t-16t^2, where [b] and [d] are constants, we can apply calculus to find the maximum height.
First, let's find the derivative of s with respect to t.
ds/dt = d([b]*[d]+64t-16t^2)/dt
Now, using the power rule of differentiation, which states that d(x^n)/dx = n*x^(n-1), the derivative becomes:
ds/dt = 0 + 0 + 64 - 32t
To find the maximum height, we set the derivative equal to zero and solve for t:
0 + 64 - 32t = 0
32t = 64
t = 64/32
t = 2
Next, let's find the second derivative of s with respect to t.
d^2s/dt^2 = d(64 - 32t)/dt
d^2s/dt^2 = -32
Now, we can determine the concavity of the function by checking the sign of the second derivative.
Since d^2s/dt^2 = -32 (< 0), the second derivative is negative, indicating that the function is concave down.
Lastly, knowing that the function is concave down and the critical point occurs at t = 2, we can conclude that this critical point is the maximum height.
Therefore, the greatest height above the ground the rock will go is when t = 2.