MATH SOLVE

3 months ago

Q:
# A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 32t + 6. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height? 1 s; 54 ft 2 s; 6 ft 2 s; 22 ft 1 s; 22 ft

Accepted Solution

A:

For this case we have the following equation:

h = -16t2 + 32t + 6

Deriving we have:

h '= -32t + 32

Equaling to zero:

-32t + 32 = 0

We clear the time:

t = 32/32

t = 1 s

We are now looking for the maximum height:

h (1) = -16 * (1) ^ 2 + 32 * (1) + 6

h (1) = 22 feet

Answer:

The ball reaches its maximum height in:

t = 1 s

The ball's maximum height is:

h (1) = 22 feet

option: 22 ft, 1 s

h = -16t2 + 32t + 6

Deriving we have:

h '= -32t + 32

Equaling to zero:

-32t + 32 = 0

We clear the time:

t = 32/32

t = 1 s

We are now looking for the maximum height:

h (1) = -16 * (1) ^ 2 + 32 * (1) + 6

h (1) = 22 feet

Answer:

The ball reaches its maximum height in:

t = 1 s

The ball's maximum height is:

h (1) = 22 feet

option: 22 ft, 1 s