Question

A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 69 m in 6 s. What is the speed (in m/s) at a height of 69 m?

A. 29.4 m/s
B. 39.2 m/s
C. 49.0 m/s
D. 58.8 m/s

Answer 1

The speed of the rocket at a height of 69 m is approximately 23.02 m/s.

Step-by-step explanation:

To find the speed of the rocket at a height of 69 m, we can use the kinematic equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

In this case, the rocket starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) can be found using another kinematic equation:

s = ut + 0.5at^2

Since the rocket reaches a height of 69 m in 6 s, we can substitute the values into the equation:

69 = 0.5 * a * (6^2)

Solving for a, we get:

a = (69 * 2) / 36 = 3.83 m/s^2

Now we can use the first equation to find the final velocity:

v^2 = (0^2) + 2 * (3.83) * 69

Simplifying, we get:

v^2 = 529.74

Taking the square root of both sides, we get:

v = 23.02 m/s