Algebra 2 math finding maximum/minimum and x intercepts

Question

The height of a rocket shot into the air is modeled by the equation h(t)=-16t^2-6t+302. Where h is the height in meters of the ticket after t seconds.

Find the maximum height of the rocket and when it occurs and when does the rocket return back to earth.

Answer 1

IF there is no negative time, then the maximum height is at time t=0 with maximum height h=302 meters.

Another strange thing about the problem as given is it looks more like it was intended to be measured in feet, not meters. -32 ft/sec/sec is the acceleration using our American system of length. 9.8 meters/sec/sec is the acceleration in the metric system. Take the derivative of the formula and you get -32t-6. The -32t represents the acceleration. The -6 represents the initial velocity and the 302 represents the initial height.

Set -32t-6=0 and solve for t=-6/32=-3/16 seconds. That's when the maximum height is, if you allow for negative time. Plug that value of t into the original quadratic equation to get the maximum height.

h(-3/16)=-16(-3/16)²-6(-3/16)+302= -9/16+18/16+302 = 9/16+302 = 302 9/16 or slightly over 302.5 meters

Set the original h(t)=-16t²-6t+302=0 and solve for t to find when it hits earth.

divide by 2 to simplify a little: -8t²-3t+151 = 0

use the quadratic formula. t= [3 + or - (9-4(-8)(151))^1/2]/(-16)

ignore the positive sign in front of the square root

t=-3/16 - (9+32(151))^1/2/(-16)

=-3/16 + 4841^1/2/16

=-3/16 + 69.58/16

=66.5816 = 4.16 seconds

check that answer by just trying t=4 seconds and plug into the original equation

-16(16)-6(4)+302 = -256-24+302= 22 meters above earth

Less than a 1/5 of a second later it hits earth

So, maximum height is about 302 meters. Time to hit earth is a little over 4 seconds