Find the maximum height ymax reached by the rocket.?
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1 m/s2 . The acceleration period lasts for time 8.00 s until the fuel is exhausted. After that, the rocket is in free fall.
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answers:
oldprof say: A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration A = 44.1 m/s2 . The acceleration period lasts for time T = 8.00 s until the fuel is exhausted.

First stage: Reaches height h = 1/2 AT^2 and speed U = AT
Second stage: Reaches max height H = h + U^2/2g = 1/2 AT^2 + (AT)^2/2g = .5*44.1*8^2 + (44.1*8)^2/(2*9.8) = 7761.6 = 7.76 km. ANS.
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electron1 say: The first two steps are to determine the vertical distance it moves in 8 seconds and to determine the rocket’s height at this time. Let’s use the following equation to determine the vertical distance it moves in 8 seconds.

d = vi * t + ½ * a * t^2, vi = 0
d = ½ * 44.1 * 8^2 = 1,411.2 meters

Let’s use the following equation to determine the rocket’s velocity at this height.

vf = vi + a * t , vi = 0
vf = 44.1 * 8 = 352.8 m/s

From now on, the rocket’s acceleration is -9.8 m/s^2. Let’s use the following equation to determine the vertical distance it moves as its velocity decreases from 352.8 m/s to 0 m/s.

vf^2 = vi^2 + 2 * g * d
0 = 352.8^2 + 2 * -9.8 * d
19.6 * d = 352.8^2
d = 124,467.84 ÷ 19.6 = 6,350.4 meters

Maximum height = 1,411.2 + 6,350.4 = 7,761.6 meters

I hope this is helpful for you.
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hfshaw say: In one dimension, the position of an object subject to a constant acceleration is given by:

x(t) = x₀ + v₀*t + a*t²/2

where x₀ and v₀ are the position and velocity of the object at t = 0.

Its velocity is given by:

v(t) = v₀ + a*t



In this case, after 8 seconds, the position of the rocket is given by:

x(8 s) = 0 + 0*t + (44.1 m/s²)*(8s)²/2 = 1411 m

and its velocity at that time is given by:

v(8 s) = 0 + (44.1 m/s²)*(8s) = 352.8 m/s

At the highest point in the rocket's trajectory, its velocity is zero (changing from positive to negative as it starts to fall under the influence of gravity). The time at which this happens once the rocket's thrust shuts off is given by:

0 m/s = 352.8 m/s - g*t_max = 352.8 m/s - (9.81 m/s²) *t_max

t_max = (352.8 m/s)/(9.81 m/s²) = 35.96 s

Now plug this into the equation for the position, using the position and velocity at the time the thrust shuts off as the initial position and velocity:

x_max = 1411 m + (352.8 m/s)*(35.96 s) - (9.81 m/s²)*(35.96 s)²/2

x_max = 7750 m
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