To solve for the time it took the rocket to reach a height of 170 meters, we can apply the kinematic equations of motion. The key equation that relates displacement ($s$), initial velocity ($u$), acceleration ($a$), and time ($t$) is:

$v = u + at$ $s = ut + \frac{1}{2} a t^2$

Where:

• $v = 197 \, \text{m/s}$ is the final velocity,
• $u$ is the initial velocity (which we do not know),
• $a = 26.3 \, \text{m/s}^2$ is the constant acceleration,
• $s = 170 \, \text{m}$ is the displacement (height),
• $t$ is the time (what we are solving for).

First, we can use the following kinematic equation that does not require time directly:

$v^2 = u^2 + 2as$

Solving for $u$ (the initial velocity):

$u^2 = v^2 - 2as$

Substituting the given values:

$u^2 = (197)^2 - 2 \times 26.3 \times 170$

Let me calculate this.We found that $u^2 = 29867$, so the initial velocity $u$ is:

$u = \sqrt{29867} \approx 172.88 \, \text{m/s}$

Now, using the first kinematic equation:

$v = u + at$

We can solve for $t$:

$t = \frac{v - u}{a}$

Substituting the values:

$t = \frac{197 - 172.88}{26.3}$

I'll calculate the value of $t$.The time it took the rocket to reach the height of 170 meters is approximately $0.92 \, \text{seconds}$.

Would you like more details or have any questions? Here are some related questions you might consider:

1. How would the time change if the acceleration were different?
2. What is the rocket’s velocity after 5 seconds of flight?
3. What would the displacement be after 3 seconds?
4. How long would it take to reach 500 meters height with the same conditions?
5. What is the significance of the initial velocity in determining the flight path?

Tip: Always check the units in kinematic problems to ensure consistency between velocity, acceleration, and time.