Final answer:
The height of the rocket after 1 second is 304 feet. The maximum height reached by the rocket is 450 feet, and it takes 3.75 seconds to reach this height. The rocket takes 8 seconds to hit the ground after launch.
Step-by-step explanation:
The question involves applying the quadratic equation for projectile motion to determine various aspects of a rocket's flight. Given the equation h(t) = -16t2 + 120t + 200, where h(t) represents the height of the rocket at time t seconds after launch:
a. To find the height of the rocket after 1 second, substitute t = 1 into the equation:
h(1) = -16(1)2+ 120(1) + 200 = 304 feet
b. The maximum height of the rocket can be found by determining the vertex of the parabola. This occurs at t = -b/(2a): tmax = -120/(2*(-16)) = 3.75 seconds
Now, substitute tmax back into h(t) to find the maximum height:
h(3.75) = -16(3.75)2 + 120(3.75) + 200 = 450 feet
c. As calculated above, the rocket takes 3.75 seconds to reach its maximum height.
d. To determine how long it takes the rocket to hit the ground, solve h(t) = 0. This requires applying the quadratic formula or factoring, if possible. Solving the equation gives two roots, one negative (which we discard as non-physical) and another positive, which represents the time the rocket takes to return to ground level: t = 8 seconds
Complete Question:
A rocket is launched from the top of a 200-foot tall cliff with an initial velocity of 120 feet per second. The height, h(t), of the rocket at time i seconds after it was launched can be modeled by the equation h(t) = -16t^2 120t+ 200
a. What was the height of the rocket after 1 second?
b. What is the maximum height of the rocket?
c. How long did it take the rocket to reach its maximum height?
d. How long does it take the rocket to hit the ground?