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?
Accepted Solution
A:
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