A ball is thrown into the air with an upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+40t+6.
A. In how many seconds does the ball reach its maximum height?
B. What's the ball's maximum height? ...?

A)
The maximum of a function can be by found equating its first derivative to 0.
dh/dt = -32t + 40
0 = -32t + 40
t = 1.25 seconds

B)
The maximum height is found by plugging the value of t from part A into the original equation:
h = -16(1.25)² + 40(1.25) + 6
h = 31 feet

Answer Link

A ball is thrown into the air with an upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+40t+6. A. In how many seconds does the ball reach its maximum height? B. What's the ball's maximum height? ...?

Respuesta :

Otras preguntas

A ball is thrown into the air with an upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+40t+6.
A. In how many seconds does the ball reach its maximum height?
B. What's the ball's maximum height? ...?