Help with this (calculus)

(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).

So the projectile has a downward velocity of 60 m/s at the moment it touches the ground.

Hello!

Find the maximum height by finding the maximum of the function, where the derivative switches from positive to negative. Find the derivative of the equation:

h(t) = 100 +40t -5t²

Power rule:

h'(t) = 40 - 10t

Set the equation equal to 0:

0 = 40 - 10t

-40 = -10t

t = 4

Test values on both sides of 4 to ensure that there is a maximum at this x value:

40 - 10(3) = 10

40 - 10(5) = -10

Therefore, the derivative switches from positive to negative. There is a maximum at t = 4. Find the height by plugging in this value for time into the original equation:

h(4) = 100 + 40(4) - 5(4)²

h(4) = 100 + 160 - 80

h(4) = 180 meters

Find the velocity when it hits the ground. Begin by finding the time necessary for the object to hit the ground:

0 = 100 + 40t - 5t²

Rearrange and factor:

0 = -5t² + 40t + 100

0 = (5t + 10)(-t + 10)

Set each factor equal to 0:

5t + 10 = 0

t = -2

-t + 10 = 0

t = 10

Therefore, the object takes 10 seconds to reach the ground. Plug this into the derivative of the equation (solved for earlier), which is the velocity:

h'(t) = 40 - 10t

h'(10) = 40 - 10(10)

h'(1) = -60 m/s. This is the *downward* velocity when the object hits the ground.

step-by-step explanation:

not sure about this one man