A rocket launched accelerates at 3.5m/s^2 in 5.90 secs and2.98m/s^2 in the next 5.98 secs and then experiences a free fall. What time will the rocket be in air?
Assume that the rocket is launched from the ground.
d1 = 0.5*3.5*5.9^2 = 60.9 m
V1=a*t = 3.5*5.9 = 20.65 m/s. = Velocity
after 3.5 s.
d2 = Vo*t + 0.5a*t^2
d2 = 20.65*5.98 + 2.98*5.98^2 = 176.8 m.
V2 = a*t = 2.98*5.98 = 17.82 m/s. =
Velocity @ 176.8 m.
d3 = (V^2-Vo^2)/2g
d3 = (0-17.82^2)/-19.6 = 16.2 m. = Free fall distance up.
t = (V-Vo)/g = (0-17.82)/-9.8 = 1.82 s.
= Time to reach max. Ht.
Tr = 5.90+5.98+1.82 = 13.7 s.
h = d1 + d2 + d3
h = 60.9 + 176.8 + 16.2 = 254 m. Above
Gnd.
h = Vo*t + 0.5g*t^2 = 254 m.
0 + 4.9t^2 = 254
t^2 = 51.8
Tf = 7.2 s. = Fall time.
T = Tr + Tf = 13.7 + 7.2 = 20.9 s. =
Time in air.
please explain clearly and precisely
d1 = 0.5*3.5*5.9^2 = 60.9 m
V1=a*t = 3.5*5.9 = 20.65 m/s. = Velocity
after 3.5 s.
d2 = Vo*t + 0.5a*t^2
d2 = 20.65*5.98 + 0.5*2.98*5.98^2 = 176.8 m.
V2 = a*t = 2.98*5.98 = 17.82 m/s. =
Velocity @ 176.8 m.
d3 = (V^2-Vo^2)/2g
d3 = (0-17.82^2)/-19.6 = 16.2 m. = Free fall distance up.
t = (V-Vo)/g = (0-17.82)/-9.8 = 1.82 s.
= Time to reach max. Ht.
Tr = 5.90+5.98+1.82 = 13.7 s.
h = d1 + d2 + d3
h = 60.9 + 176.8 + 16.2 = 254 m. Above
Gnd.
h = Vo*t + 0.5g*t^2 = 254 m.
0 + 4.9t^2 = 254
t^2 = 51.8
Tf = 7.2 s. = Fall time.
T = Tr + Tf = 13.7 + 7.2 = 20.9 s. =
Time in air.
Well, let's break it down. In the first 5.90 seconds, the rocket accelerates at 3.5m/s^2. In the next 5.98 seconds, it accelerates at 2.98m/s^2. So the rocket accelerates for a total of 5.90 + 5.98 = 11.88 seconds.
After that, the rocket experiences free fall, which means it's no longer accelerating. So we can say that the total time the rocket is in the air is 11.88 seconds.
But hey, don't worry, the rocket won't be alone in the air. It'll have gravity to keep it company!
To determine the total time the rocket will be in the air, we first need to calculate the time it takes for the rocket to reach its maximum acceleration in each phase.
Phase 1: Accelerating at 3.5 m/s^2 for 5.90 seconds.
Using the equation of motion:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
Since the rocket is launched from rest, the initial velocity (vi) is 0 m/s. So, we can calculate the final velocity (vf) in phase 1:
vf = vi + at
vf = 0 + (3.5 m/s^2)(5.90 s)
vf = 20.65 m/s
Now, we can calculate the time it takes for the rocket to reach this velocity in phase 1:
vf = vi + at
20.65 m/s = 0 + (3.5 m/s^2)t1
t1 = 20.65 m/s / 3.5 m/s^2
t1 ≈ 5.90 seconds
Phase 2: Accelerating at 2.98 m/s^2 for 5.98 seconds.
Using the same equation, we can calculate the final velocity (vf) in phase 2:
vf = vi + at
vf = 20.65 m/s + (2.98 m/s^2)(5.98 s)
vf ≈ 38.139 m/s
Now, let's calculate the time it takes for the rocket to reach this velocity in phase 2:
vf = vi + at
38.139 m/s = 20.65 m/s + (2.98 m/s^2)t2
t2 = (38.139 m/s - 20.65 m/s) / (2.98 m/s^2)
t2 ≈ 5.98 seconds
The total time the rocket is in the air is the sum of the times from both phases:
Total time = t1 + t2
Total time ≈ 5.90 s + 5.98 s
Total time ≈ 11.88 seconds
Therefore, the rocket will be in the air for approximately 11.88 seconds.