A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?
To find the maximum elevation reached by the rocket, we can use the kinematic equation that relates displacement (Δy), initial velocity (vi), time (t), and acceleration (a):
Δy = vi * t + 0.5 * a * t^2
In this case, the rocket starts from rest, so the initial velocity (vi) is 0 m/s. The acceleration (a) is given as 20 m/s^2 upward, and the time (t) is 4.0 seconds.
Let's plug in the values into the equation:
Δy = 0 * 4.0 + 0.5 * 20 * (4.0)^2
Simplifying further:
Δy = 0 + 0.5 * 20 * 16
Δy = 0 + 160
Δy = 160 meters
Therefore, the maximum elevation reached by the rocket is 160 meters.
To determine the maximum elevation reached by the rocket, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = displacement or elevation
u = initial velocity (since the rocket starts from rest, u = 0)
a = acceleration
t = time
In this case, the acceleration is positive 20 m/s^2 (upward direction) and the time is 4.0 seconds. Plugging these values into the equation, we get:
s = 0 + (1/2)(20)(4)^2
s = 0 + 10(16)
s = 160 meters
Therefore, the maximum elevation reached by the rocket is 160 meters.
well, do the power thing first
v = 0 + a t
v = Vi = 20 * 4 = 80 m/s at power cutoff
at what height did the power stop?
h = 0 + 0 t + (1/2)(20)(16)
= 160 meters
so
we have Vi = 80 m/s at Hi = 160 meters
now coast up
v = Vi - g t
at top v = 0 , using g = 9.8
0 = 80 - 9.8 t
t = 8.16 seconds coasting to peak
then
h = Hi + Vi t - 4.9 t^2
h = 160 + 80 (8.16 = 4.9(8.16)^2