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