Answer:
(a)0.625s (b)1.569s
Explanation:
a.The ball reaches its maximum height when its speed = 0, or changing from positive to negative. To find out the time t for this we need to get the velocity function by taking the first derivative of the height function:
[tex]v(t) = s^{'}(t) = (-16t^2)^{'} + (20t)^{'} + 8^{'} = -32t + 20[/tex]
So when v(t) = 0
[tex]-32t + 20 = 0[/tex]
[tex]t = 20/32 = 0.625s[/tex]
b. The ball land back on the ground when s(t) = 0:
[tex]-16t^2 + 20t + 8 = 0[/tex]
[tex]4t^2 - 5t - 2 = 0[/tex]
[tex]t^2 - \frac{5}{4}t + \frac{1}{2} = 0 [/tex]
[tex]t \approx 1.569s[/tex]
