Question

A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 52.7 m/s2 for 1.41 s. its fuel is then exhausted, so it continues upward as a free-fall particle and then falls back down. (a) what is the maximum altitude reached? (b) what is the total time elapsed from takeoff until the rocket strikes the ground?

Answer 1

The maximum altitude reached by the rocket is 334.2 meters, and the total time elapsed from takeoff until the rocket strikes the ground is 16.55 seconds.

Step-by-step explanation:

To find the maximum altitude reached by the rocket, we need to consider two stages: the powered ascent and the free-fall descent. During the powered ascent, the rocket accelerates upwards at a constant acceleration of 52.7 m/s2 for 1.41 seconds.

Using the kinematic equation for displacement: s = ut + (1/2)at2, where 's' is displacement, 'u' is initial velocity (0 m/s in this case, as it starts from rest), 'a' is acceleration, and 't' is time, we get:

s = 0 m/s * 1.41 s + 0.5 * 52.7 m/s2 * (1.41 s)2 = 52.3 meters

Now, the velocity at the end of the powered ascent can be found using the equation v = u + at, giving us v = 0 m/s + 52.7 m/s2 * 1.41 s = 74.3 m/s. This is the initial velocity for the free-fall ascent.

For the free-fall, the only acceleration is due to gravity, which is -9.81 m/s2 (negative as it opposes the initial velocity). We use v2 = u2 + 2as to find the additional altitude gained in free-fall when the velocity reaches zero (v = 0 m/s at the maximum height). Solving for 's', we get:

s = (0 m/s)2 - (74.3 m/s)2 / (2 * -9.81 m/s2) = 281.9 meters

So, the total maximum altitude is the sum of the displacement during the powered ascent and the additional altitude gained in free-fall which is 52.3 meters + 281.9 meters = 334.2 meters.

To find the total time elapsed until the rocket strikes the ground, we'll add the time of ascent, the time to reach maximum altitude during free-fall, and the time it takes to fall back down from that altitude. We've already found the time of ascent (1.41 s) and can use the same velocity and acceleration to find the time to reach maximum altitude during free-fall with the formula t = (v - u) / a, giving us t = (0 m/s - 74.3 m/s) / -9.81 m/s2 = 7.57 s. For the descent, the time from maximum altitude to the ground will be equal to twice the time from the end of powered ascent to maximum altitude, since the paths are the same and the final velocities (at the ground) will be equal in magnitude but opposite in direction.

The total time elapsed from takeoff until the rocket strikes the ground is then 1.41 s + 2 * 7.57 s = 16.55 seconds.

Answer 2

For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is

a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.

-52.7 = (0-v1)/1.41
v1 = 74.31 m/s

We can use v1 for the formula of the maximum height attained by an object thrown upwards:

Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m

The maximum height attained by the model rocket is 281.42 m.

For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.

Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.

2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)

Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s

Therefore, the total time= 1.41+6.83+7.57 = 15.81 s