A test rocket is fired up from rest. The net acceleration is 20m/s^2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?
d = 1/2 (a)(t)^2 = 160 m
at rest, Vo = u = 0 m/s
I found the final velocity by using
v = u +at = 0 + (20)(4) = 80 m/s
d = vt
t = 160 m / 80 m/s = 2 s
I might be wrong at this
I know d= Vot - 1/2 (a)(t)^2
By any chance, what textbook are you using?
h=at²/2=20•16/2=160 m
To find the maximum elevation reached by the rocket, we can use the equations of motion.
The initial velocity of the rocket, u, is 0 m/s as it is fired up from rest.
The acceleration, a, is 20 m/s^2 upward, but we can consider it as -20 m/s^2 (negative sign denotes the opposite direction of the upward motion).
The time taken, t, is 4.0 seconds.
We need to find the maximum elevation (height), which we'll call H.
The equation that relates height, initial velocity, time, and acceleration is given by:
H = ut + (1/2)at^2
Substituting the given values:
H = 0 * 4.0 + (1/2)(-20)(4.0)^2
H = 0 + (1/2)(-20)(16)
H = 0 + (-10)(16)
H = -160 meters
The negative sign indicates that the rocket reached a maximum elevation of 160 meters below its starting point, i.e., it went in a downward direction.