The flight of one type of rocket is described by a function h = −5(t − 3)2 + 50, where
h is the height of the rocket, in metres, and t is the time, in seconds, since the rocket
was fired.
a. What was the max height of the rocket?
b. At what time the max height was reached?
c. What was the initial height of the rocket?
d. What was the height of the rocket after 2 seconds of the flight? At what another
time this height was reached?
e. Was the rocket in the air after 6 sec?
To find the answers, let's break down the given function and solve for each question:
a. The maximum height of the rocket can be found by looking at the vertex of the quadratic function. The vertex form of a quadratic function is given as h = a(t - h)^2 + k, where (h, k) represents the vertex. In this case, the given function is in vertex form with a = -5, h = 3, and k = 50. The vertex can be found using the formula h = -b/(2a). Substituting the values, we have h = -(-5)/(2*-5) = 5/10 = 0.5. Therefore, the maximum height of the rocket is 0.5 meters.
b. The time at which the maximum height is reached can be found by substituting the value of h into the original equation and solving for t. Substituting h = 0.5 into the equation h = -5(t - 3)^2 + 50, we get 0.5 = -5(t - 3)^2 + 50. Simplifying, we have -0.5 = -5(t - 3)^2. Dividing by -5, we get (t - 3)^2 = 0.1. Taking the square root of both sides, we have t - 3 = ±√0.1. Solving for t, we have t = 3 ± √0.1. Therefore, the time at which the maximum height is reached is approximately t = 3 ± 0.3162 seconds.
c. The initial height of the rocket can be found by substituting t = 0 into the given function h = -5(t - 3)^2 + 50. Simplifying, we have h = -5(0 - 3)^2 + 50 = -5(9) + 50 = -45 + 50 = 5. Therefore, the initial height of the rocket is 5 meters.
d. To find the height of the rocket after 2 seconds of flight, substitute t = 2 into the given function h = -5(t - 3)^2 + 50. Simplifying, we have h = -5(2 - 3)^2 + 50 = -5(-1)^2 + 50 = -5(1) + 50 = -5 + 50 = 45 meters. The same height of 45 meters is reached again when t = 4 seconds.
e. To determine whether the rocket is in the air after 6 seconds, we need to check if the height is greater than zero. Substituting t = 6 into the given function h = -5(t - 3)^2 + 50, we have h = -5(6 - 3)^2 + 50 = -5(3)^2 + 50 = -5(9) + 50 = -45 + 50 = 5 meters. Since the height is 5 meters, which is greater than zero, the rocket is still in the air after 6 seconds.
In summary,
a. The max height of the rocket is 0.5 meters.
b. The time at which the max height was reached is approximately t = 3 ± 0.3162 seconds.
c. The initial height of the rocket is 5 meters.
d. The height of the rocket after 2 seconds is 45 meters. The same height of 45 meters is reached again when t = 4 seconds.
e. Yes, the rocket is still in the air after 6 seconds.