Question

A ball is thrown directly upward from a height of 8 ft with an initial velocity of 20 ​ft/sec. The function ​s(t) = −16t²+20t+8 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. a. Determine the time at which the ball reaches its maximum height and find the maximum height.b. How long after the ball is thrown does it land back on the ground?

Answer 1

(a)0.625s (b)1.569s

Step-by-step 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:

`v(t) = s^(')(t) = (-16t^2)^(') + (20t)^(') + 8^(') = -32t + 20`

So when v(t) = 0

`-32t + 20 = 0`

`t = 20/32 = 0.625s`

b. The ball land back on the ground when s(t) = 0:

`-16t^2 + 20t + 8 = 0`

`4t^2 - 5t - 2 = 0`

`t^2 - (5)/(4)t + (1)/(2) = 0`

`t \approx 1.569s`