Final answer:
To describe the motion of a rocket launched from a height of 3 m with an initial velocity of 15 m/s, the kinematic equation for projectile motion in the vertical direction would be used, incorporating gravity's acceleration.
Step-by-step explanation:
To write an equation for a rocket launched from a height of 3 m with an initial velocity of 15 m/s, one would use the kinematic equation for projectile motion. Assuming the acceleration due to gravity is -9.81 m/s2 (negative because it is directed downwards), we can use the following equation for the vertical motion (y-direction):
y(t) = y0 + v0t + ½at2
Where:
y(t) is the height of the rocket at time t.
y0 is the initial height, which is 3 m.
v0 is the initial velocity, which is 15 m/s vertically upwards.
a is the acceleration, which is -9.81 m/s2 due to gravity.
t is the time in seconds.
The full equation for this specific launch would be:
y(t) = 3 + 15t - 4.905t2
Note that the horizontal motion is not affected by gravity and would be simply described by x(t) = v_horizontal * t, where v_horizontal is the horizontal component of the initial velocity if any. Since the initial velocity is given as 15 m/s upwards and no horizontal component is mentioned, we assume there is no horizontal motion and focus only on the vertical motion.