(a) h(t) is quadratic, so we can complete the square to write
h(t) = 100 + 40 t - 5 t²
h(t) = 100 - 5 (t² - 8 t)
h(t) = 100 - 5 (t² - 8 t + 16 - 16)
h(t) = 100 + 80 - 5 (t² - 8 t + 16)
h(t) = 180 - 5 (t - 4)²
The squared quantity is always non-negative, which means h(t) has a maximum value of 180 m.
(b) First, find when the projectile hits the ground, which happens when h(t) = 0:
0 = 180 - 5 (t - 4)²
180 = 5 (t - 4)²
36 = (t - 4)²
± 6 = t - 4
t = -2 OR t = 10
Omit the negative solution. The velocity of the ball at t = 10 s is equal to the derivative at the point, h' (10).
[tex]h'(10)=\displaystyle\lim_{t\to10}\frac{h(t)-h(10)}{t-10}[/tex]
[tex]h'(10)=\displaystyle\lim_{t\to10}\frac{(180-5(t-4)^2)-80}{t-10}[/tex]
[tex]h'(10)=\displaystyle\lim_{t\to10}\frac{100+40t-5t^2}{t-10}[/tex]
[tex]h'(10)=\displaystyle-5\lim_{t\to10}\frac{(t+2)(t-10)}{t-10}[/tex]
[tex]h'(10)=\displaystyle-5\lim_{t\to10}(t+2)=-60[/tex]
So the projectile has a downward velocity of 60 m/s at the moment it touches the ground.
