A rocket has mass 0.8kg and launches straight upwards with initial velocity 6 m/s. This time the thrusting force decreases with time as the rocket runs out of fuel, like F(t) = 22-7t Newtons. How many seconds after launch does the rocket reach its peak height?

To determine when the rocket reaches its peak height, we need to find when its velocity becomes zero. The peak height is reached when the upward velocity slows to a stop.

We are given:

- Initial mass of the rocket \( m = 0.8 \, \text{kg} \),
- Initial velocity \( v_0 = 6 \, \text{m/s} \),
- Thrusting force as a function of time: \( F(t) = 22 - 7t \, \text{N} \),
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \).

### Step 1: Find the net force on the rocket
The net force on the rocket is the thrusting force minus the gravitational force. The gravitational force is constant and equal to \( F_g = mg = 0.8 \times 9.8 = 7.84 \, \text{N} \).

Thus, the net force as a function of time is:
\[
F_{\text{net}}(t) = F(t) - F_g = (22 - 7t) - 7.84 = 14.16 - 7t \, \text{N}.
\]

### Step 2: Use Newton's second law
Newton's second law relates the net force to acceleration:
\[
F_{\text{net}}(t) = ma(t),
\]
where \( a(t) \) is the acceleration. We can solve for acceleration:
\[
a(t) = \frac{F_{\text{net}}(t)}{m} = \frac{14.16 - 7t}{0.8}.
\]
Simplifying this expression:
\[
a(t) = 17.7 - 8.75t \, \text{m/s}^2.
\]

### Step 3: Integrate acceleration to find velocity
To find the velocity, we integrate the acceleration:
\[
v(t) = \int a(t) \, dt = \int (17.7 - 8.75t) \, dt = 17.7t - 4.375t^2 + C.
\]
The constant \( C \) can be found from the initial condition: \( v(0) = 6 \, \text{m/s} \). Substituting this into the velocity equation:
\[
v(0) = 17.7 \times 0 - 4.375 \times 0^2 + C = 6,
\]
so \( C = 6 \).

Thus, the velocity as a function of time is:
\[
v(t) = 17.7t - 4.375t^2 + 6.
\]

### Step 4: Solve for when the velocity is zero
At the peak height, the velocity is zero:
\[
0 = 17.7t - 4.375t^2 + 6.
\]
Rearranging this equation:
\[
4.375t^2 - 17.7t - 6 = 0.
\]
This is a quadratic equation. We can solve it using the quadratic formula:
\[
t = \frac{-(-17.7) \pm \sqrt{(-17.7)^2 - 4(4.375)(-6)}}{2(4.375)}.
\]
Simplifying inside the square root:
\[
t = \frac{17.7 \pm \sqrt{313.29 + 105}}{8.75} = \frac{17.7 \pm \sqrt{418.29}}{8.75}.
\]
Taking the square root:
\[
t = \frac{17.7 \pm 20.45}{8.75}.
\]
Thus, we have two solutions:
\[
t_1 = \frac{17.7 + 20.45}{8.75} = \frac{38.15}{8.75} \approx 4.36 \, \text{seconds},
\]
\[
t_2 = \frac{17.7 - 20.45}{8.75} = \frac{-2.75}{8.75} \approx -0.31 \, \text{seconds} \, (\text{not physically meaningful}).
\]

### Conclusion:
The rocket reaches its peak height at approximately \( t = 4.36 \, \text{seconds} \).

Would you like further details or have any questions?

Here are 5 related questions to expand on this:
1. What is the maximum height reached by the rocket?
2. How would the problem change if air resistance is considered?
3. How does the mass of the rocket affect the time to reach peak height?
4. How long will it take for the rocket to return to the ground?
5. How would you calculate the velocity at any time \( t \) during ascent?

**Tip**: Always check units for consistency when solving physics problems.

Learn how to calculate the time for a rocket to reach its peak height with an initial velocity of 6 m/s and a decreasing thrust force, using Newton's second law and quadratic equations. This step-by-step solution explores key physics concepts, including force, acceleration, and velocity. Suitable for advanced high school physics students.

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