I posted this question earlier and thank you for respnding, but the answer you give me was incorrect. I try looking at hw you solve and cant find the mistake but when when I check your answer it said it was wrong. Can you please look over it again, I greatly appreciate it

a toy rocket moving vertically upward passes by a 2.2m- high window whose sill is 7.0m above the ground. the rocket takes 0.17s to travel the 2.2m height of the wondow
part a): what was the launch speed of the rocket ? assume the propellant is burned very quickly at blastoff

part b) how high will the rocket go ?

What is the correct answer? I'll try to understand where is the mistake.

I‘ve found the mistake

h(o) = v(o)•t - g•t^2/2,
2 h(o) = 2 v(o)•t - g•t^2
v(o) = h(o) /t +gt/2 = 2.2/0.17 + 9.8•0.17/2 = 13.77 m/s.
Then try to calculate yourself…

I apologize for any confusion earlier. Let's go through the problem once again to find the correct answers.

In this problem, we need to find the launch speed of the rocket (Part a) and the maximum height it reaches (Part b).

For Part a), we can use the equation of motion:

v = u + at

Where:
v = final velocity (which is 0 since the rocket reaches its maximum height and its velocity becomes zero)
u = initial velocity (which is what we're trying to find)
a = acceleration
t = time taken to travel the 2.2m height

Since the rocket is moving vertically upward and slowing down, we can assume the acceleration due to gravity acting on it is -9.8 m/s^2. The negative sign indicates the opposite direction to the motion.

Using the equation above, we can rearrange it to solve for u:

u = v - at

Plugging in the values:
v = 0 m/s (final velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
t = 0.17 s (time taken to travel 2.2m)

u = 0 - (-9.8 * 0.17)
u = 1.666 m/s (rounded to three decimal places)

Therefore, the launch speed of the rocket is approximately 1.666 m/s.

Moving on to Part b), we can use a different equation to find the maximum height reached by the rocket.

h = u*t + (1/2)*a*t^2

Where:
h = maximum height reached by the rocket (which we're trying to find)
u = initial velocity (1.666 m/s)
t = total time taken (0.17 s as given)
a = acceleration due to gravity (-9.8 m/s^2)

Plugging in the values:
h = (1.666 * 0.17) + (1/2) * (-9.8) * (0.17)^2
h = 0.283 + (-0.141)
h = 0.142 m (rounded to three decimal places)

Therefore, the rocket reaches a maximum height of approximately 0.142 m above the window sill.

Please double-check these calculations and let me know if there are any further questions or issues.