A model rocket is launched straight upward
with an initial speed of 51.6 m/s. It acceler-
ates with a constant upward acceleration of 1.96 m/s2 until its engines stop at an altitude of 230 m. The acceleration of gravity is 9.81m/s^2.
When does the rocket reach maximumheight?
Answer in units of s
How long is the rocket in the air?
Answer in units of s
Lets look at the velocity at 230m
Vf^2=Vi^2+2a*230
solve for Vf (at 230)
vavg= (Vf+51.6)/2
time to go first 230m= 230/vavg
Now time to top...
Vf(top)=Vf(at 230)-gt solve for t
0=Vf(230)-9.81t
so now add this time to time to go first 230. That is the time to get to the max height.
Now , figure the time to fall from the top height. What is the max height?
initial KE+ workby rocket=finaPE
1/2 m 51.6^2+Force*230=mg H
where force= m*a solve for H.
time to fall from H: H=1/2 g t^2 solve for t.
thanks
To find the time when the rocket reaches maximum height, we can use the equation of motion for vertical motion:
v = u + at
Where:
v = final velocity (when the rocket reaches maximum height, the vertical velocity will be zero)
u = initial velocity (51.6 m/s in this case)
a = acceleration (1.96 m/s^2 in this case)
t = time
Since the final velocity is zero, we can rewrite the equation as:
0 = 51.6 + 1.96t
Solving for t gives us:
t = -51.6 / 1.96
t ≈ -26.33 s
However, since we're dealing with time, the negative value doesn't make sense in this context. Therefore, we can ignore the negative value and consider the rocket reaching its maximum height at t ≈ 26.33 s.
To find how long the rocket is in the air, we need to consider the time it took for the rocket to reach its maximum height and then come back down to the ground.
Since the rocket reaches its maximum height at t ≈ 26.33 s and the initial speed is 51.6 m/s, we can use the equation:
h = ut + (1/2)at^2
Where:
h = height (230 m in this case)
u = initial velocity (51.6 m/s in this case)
a = acceleration (1.96 m/s^2 in this case)
t = time
Rearranging the equation to solve for t gives us:
t = sqrt((2h) / a)
Plugging in the values, we have:
t ≈ sqrt((2 * 230) / 1.96)
t ≈ sqrt(460 / 1.96)
t ≈ sqrt(234.69)
t ≈ 15.31 s
Therefore, the rocket is in the air for approximately 15.31 seconds.