Using h(t)=gt^2+v0t+h0, where g= -16, h0= initial height, v0= initial velocity, and t= time
A rocket is fired into the air from the top of a building that is 200 feet tall. The initial velocity of the rocket is 600 feet/sec. If necessary, round answers to the nearest hundredth of a unit
a) Write the height function
b) When does the rocket reach its maximum height?
c) What is the rocket's maximum height?
d) When does it hit the ground?
did you sub in the given data into the general formula to get
h(t) = -16t^2 + 600t + 200 ?
for b) and c), find the vertex of the parabola using the method that was taught to you
d) set -16t^2 + 600t + 200 = 0
and solve for t
a) The height function for the rocket's trajectory can be written as:
h(t) = -16t^2 + v0t + h0
Given that g = -16 (acceleration due to gravity), h0 = 200 (initial height of the building), v0 = 600 (initial velocity of the rocket), and t represents time.
Substituting these values into the height function:
h(t) = -16t^2 + 600t + 200
b) To find when the rocket reaches its maximum height, we need to determine the time at which the rocket's vertical velocity becomes zero.
The vertical velocity of the rocket can be obtained by differentiating the height function with respect to time:
v(t) = d/dt[h(t)] = -32t + 600
Setting v(t) = 0 and solving for t:
-32t + 600 = 0
32t = 600
t = 600/32
t ≈ 18.75
Therefore, the rocket reaches its maximum height at approximately 18.75 seconds.
c) To find the rocket's maximum height, we substitute the time (t) into the height function:
h(18.75) = -16(18.75)^2 + 600(18.75) + 200
Evaluating this expression will give us the maximum height.
d) To determine when the rocket hits the ground, we need to find the time at which the height (h) becomes zero.
Setting h(t) = 0 and solving for t:
-16t^2 + 600t + 200 = 0
We can use the quadratic formula to solve for t.