+244 A new SpaceX rocket is being tested and launches from the ground. The first stage rocket engine provides constant upward acceleration during the burn phase. After the first stage engine burns out, the rocket has risen to 105.0 m and acquired an upward velocity of 50.0 m/s. The second stage fails to fire. The rocket continues to rise, reaches maximum height, and falls back to the ground. Neglect air resistance. The maximum height reached by the rocket is:
A.256 m.
B.221 m.
C.244 m.
D.233 m.
E.209 m.
+37146 We'll use a couple of different equations regading position and velocity
First: v = vo + at vo= 50 a = -9.8 m/s^2 FInd the point where the rocket reaches its apex, v=0
0 = 50 -9.8t results in t = 5.1 deconds
How high did it go?
x = xo + vo t + 1/2 a t^2 xo= 105 m vo = 50 m/s and a = -9.8 m/s^2
= 105 + 50 (5.1) - 1/2(9.8) (5.1)^2 = 360-127.45 = 232.55 m
+244 Thank You So Much:) my nest quetsion consist of: A new SpaceX rocket is being tested and launches from the ground. The first stage rocket engine provides constant upward acceleration during the burn phase. After the first stage engine burns out, the rocket has risen to 105.0 m and acquired an upward velocity of The second stage fails to fire. The rocket continues to rise, reaches maximum height, and falls back to the ground. Neglect air resistance. The time interval during which the rocket engine provided the upward acceleration, is closest to
3.0 s.
4.2 s.
1.5 s.
1.7 s.
4.0 s.
+37146
You left off the velocity the rocket has acheived.....I assume it is 50 m/s again
Velocity changed from 0 to 50 m/s in t time
v= vo + at
50 = 0 + at
50/t = a (Equation 1)
Now we know x = xo + vo t + 1/2 a t^2 for the thrust portion of the rocket flight (xo and vo are both ZERO) so we are left with
x= 1/2 a t^2 From Equation 1 : 50/t = a Substitute this value for a
105 = 1/2 (50/t) t^2
105 = 25 t
t = 4.2 seconds
