Question

Help with this (calculus)

Answer 1

(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).

`h'(10)=\displaystyle\lim_(t\to10)(h(t)-h(10))/(t-10)`

`h'(10)=\displaystyle\lim_(t\to10)((180-5(t-4)^2)-80)/(t-10)`

`h'(10)=\displaystyle\lim_(t\to10)(100+40t-5t^2)/(t-10)`

`h'(10)=\displaystyle-5\lim_(t\to10)((t+2)(t-10))/(t-10)`

`h'(10)=\displaystyle-5\lim_(t\to10)(t+2)=-60`

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

Answer 2

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.