Answer:
5.5 m/s²
Explanation:
The rocket has initial velocity 0 m/s, and acceleration a. After 4 seconds, it has a velocity of v and position h.
When the bolt falls off, it has initial velocity v and acceleration -9.8 m/s². After 6 seconds, it falls a distance of h.
For the rocket:
Δy = ½ (v + v₀) t
h = ½ (v + 0) (4)
h = 2v
For the bolt:
Δy = v₀ t + ½ at²
-h = v (6) + ½ (-9.8) (6)²
h = 176.4 − 6v
Substitute:
2v = 176.4 − 6v
8v = 176.4
v = 22.05
Therefore, the rocket's acceleration is:
a = (v − v₀) / t
a = (22.05 − 0) / 4
a = 5.5 m/s²
The rocket's acceleration is 5.5m/s^2
Data;
[tex]v=at\\t=4\\v=4a[/tex]
Using kinematic equation,
[tex]Y= V_o+V_ot+\frac{1}{2}at^2\\ y=0+0+\frac{1}{2}a(4)^2\\ y=8a[/tex]
The time it took the bolt to hit ground is 6s
[tex]Y=V_o+V_ot-\frac{1}{2}gt\\ 0=8a+4a(6)-\frac{1}{2}*9.8*6^2\\ 32a=176.4\\a=5.5m/s^2[/tex]
The rocket's acceleration is 5.5m/s^2
Learn more on kinematic equation here;
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