Final answer:
Using the kinematic equations for uniformly accelerated motion, the acceleration of the model rocket is determined to be 15 m/s^2. Subsequently, the speed at a height of 120 m is calculated to be 60 m/s.
Step-by-step explanation:
The student's question involves determining the speed of a model rocket that accelerates upwards from the ground at a constant acceleration, reaching a height of 120 m in 4 seconds. To solve this, we can use one of the kinematic equations for uniformly accelerated motion:
v = u + at
However, we need to determine the acceleration (a) first. We can use another kinematic equation to find the acceleration:
s = ut + \(\frac{1}{2}\)at^2 where s is the displacement, u is the initial speed (0 m/s, since the rocket starts from rest), t is the time, and a is the acceleration.
Substituting the given values:
120 m = 0 m/s (4 s) + \(\frac{1}{2}\)a(4 s)^2
Solving this gives us:
a = \(\frac{2s}{t^2}\) = \(\frac{2*120 m}{(4 s)^2}\) = 15 m/s^2
Now we can use the first equation to determine the velocity at 120 m:
v = 0 m/s + (15 m/s^2)(4 s) = 60 m/s
Therefore, the speed of the model rocket at a height of 120 m is 60 m/s.