A rocket moves upward, starting from rest with an acceleration of 34.4 m/s2 for 4.40 s. It runs out of fuel at the end of the 4.40 s but does not stop. How high does it rise above the ground?
after 4.40s, the height is
34.4*4-4.9*4.4^2 = 42.74m
At that point, its velocity is (34.4-9.8)*4.4 = 108.24m/s
So, now we have the ballistic trajectory of the height
42.7 + 108.24t - 4.9t^2
which has its maximum value at t=11.04
Now just plug that in to find the max height.
Steve, think you're a bit off on the height at the end of the burn. SB 331m. And I would subtract out g, the problem gives a flat 34.4 so v at that time is 151m/s.
In the second stage v goes from 151 to zero under influence of gravity so y = v^2/2a = 151^2/9.8*2 = 1169m. All told 1500m.
Sorry would NOT subtract out g in part 1
To find the height the rocket rises above the ground, we can use the kinematic equation:
\(h = h_0 + v_0t + \frac{1}{2}at^2\)
where:
- \(h\) is the final height above the ground
- \(h_0\) is the initial height (assumed to be zero in this case since the rocket starts from the ground)
- \(v_0\) is the initial velocity (also assumed to be zero because the rocket starts from rest)
- \(t\) is the time
- \(a\) is the acceleration
Given that the rocket starts from rest (\(v_0 = 0\)), the initial height is zero (\(h_0 = 0\)), and the acceleration is 34.4 m/s², we can rearrange the equation and solve for \(h\):
\(h = \frac{1}{2}at^2\)
Substituting the values:
\(h = \frac{1}{2} \times 34.4 \, \text{m/s}^2 \times (4.40 \, \text{s})^2\)
Calculating the equation:
\(h = \frac{1}{2} \times 34.4 \, \text{m/s}^2 \times 19.36 \, \text{s}^2\)
\(h = 333.824 \, \text{m}\)
Therefore, the rocket rises approximately 333.824 meters above the ground.