If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=32+64t−16t2.
What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?

a) , b)

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

Second Derivative

(absolute maximum)

The maximum height reached by the ball is:

b) The time required by the ball to hit the ground is:

Just one root offers a solution that is physically reasonable:

The velocity of the ball when it hits the ground is:

a) 96ft

b) -78.4ft/s

Explanation:

Given the height of the ball modelled as s(t)=32+64t−16t²

a) at maximum height, the velocity of the ball is zero

Velocity = d(s(t))/dt = -32t+64

Since v = 0m/s at maximum height

0 = -32t+64

32t = 64

t = 64/32

t = 2seconds

This means it takes 2 seconds for the ball to reach its maximum height.

The maximum height reached will occur at t = 2secs

s(2) =32+64(2)-16(2)²

s(2) = 32+128-64

s(2) = 96ft

b) When the ball hits the ground the height = 0

s(t) = 0

-16t²+64t+32 = 0

t²-4t-2 = 0

Factorizing the expression:

-(-4)±√(-4)²-4(1)(-2)/2(1)

= 4±√16+8/2

= 4±√24/2

= (4+2√6)/2 and (4-2√6)/2

= 2+√6 and 2-√6

t= 4.45s and -0.45s

Substitute t = 4.45s into the equation for calculating the velocity

V(t) = -32t+64

V(4.45) = -32(4.45)+64

V(4.45) = -142.4+64

V(4.45) = -78.4ft/s

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles