A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h (t) = - 16t2 + 40ft + 1.5. • About how many second after launch would the ball hit the ground? • What is the maximum height of the ball?

For this case we have the following equation:

h (t) = - 16t2 + 40t + 1.5

To hit the ground we have:

-16t2 + 40t + 1.5 = 0

Solving the polynomial we have:

t1 = - 0.04

t2 = 2.54

Taking the positive root (for being time) we have:

t = 2.54 s

For the maximum height we have:

We derive the equation:

h ' (t) = - 32t + 40

We equal zero and clear t:

t = 40/32

t = 1.25

We evaluated t = 1.25 in the function:

h (1.25) = - 16 * (1.25) ^ 2 + 40 * (1.25) + 1.5

h (1.25) = 26.5 feet

Answer:

the ball hit the ground at:

t = 2.54 s

The maximum height of the ball is:

h (1.25) = 26.5 feet