The ball will be fastest right before it hits the ground because of gravity.
First, let's find the value of t where the ball hits the ground.
-8t²+25 = 0
-8t² = -25
t² = [tex] \frac{25}{8} [/tex]
t = [tex]\frac{5}{2\sqrt{2}} [/tex]
Note: t has to be positive, which is why there is no positive/negative symbol.
Now, let's calculate the derivative of this function. The derivative will give us the speed of the ball for time t.
d/dx = -16t
Now, let's plug in the value of t
-16 ( [tex]\frac{5}{2\sqrt{2}} [/tex] }
= [tex]\frac{-5(8)}{\sqrt{2}} [/tex]
= [tex] -20 \sqrt{2} [/tex]
Take the absolute value of that because speed can't be negative
[tex]20\sqrt{2} [/tex]
That's the speed of the ball. You can convert that into decimal if you'd like.
Have an awesome day! :)
