A cannonball is launched

Question

A cannon ball is launched into the air with an upward velocity of 233 feet per second, from a 2-foot tall cannon. The height h of the cannon ball after t seconds can be found using the equation h=−16t2+233t+2.Approximately how long will it take for the cannon ball to be 328 feet high? Round answers to the nearest tenth if necessary. How long long will it take to hit the ground?

Please explain each step, thank you!

Answer 1

set h=328 and solve for t

for when it hits the ground, set h=0 and solve for t

328 = -16t^2 + 233t +2

16t^2 -233t + 326 =0

use the quadratic formula to solve for t. You usually 2 solutions.

t= 233/32 + or - (1/32)sqr(233^2 -4(16)(326) = 7.28 + or - 6.92= 14.2 or 0.36 seconds to reach 328 feet

max height is when the derivative = 0

32t-233 =0

t =233/32 = 7.28 seconds

h(7.28) = -16(7.28)^2 + 233(7.28) + 2 = 848 feet. It reaches 328 feet twice, once on the way up and then again on the way back down

for when it hits the ground set h(t)=0 and solve for t

0 = -16t^2 + 233t + 2

16t^2 -233t -2 = 0

t = 233/32 + (1/32)sqr(233^2 +128) = 7.28 + 7.29 = 14.57 seconds