es to a stop at its highest point. The time to reach the maximum height can be determined by using the kinematic equation:

$$v = u + at$$

where:

$$v$$ is the final velocity (0 m/s at the maximum height),

$$u$$ is the initial velocity,

$$a$$ is the acceleration (which is $$-g$$ due to gravity), and

$$t$$ is the time.

At the maximum height:

$$0 = u - gt$$

Solving for $$t$$ gives us:

$$t = \frac{u}{g}$$

The given problem states that the time interval between the throws is 3 seconds. Therefore, this time must equal the time taken to reach the maximum height. Thus:

$$t = 3 \text{ seconds}$$

Using the above formula:

$$t = \frac{u}{g} = 3$$

Solving for $$u$$, we get:

$$u = 3g = 3 \times 10 = 30 \text{ m/s}$$

Now, to find the maximum height, we use another kinematic equation:

$$h = ut - \frac{1}{2}gt^2$$

Substituting the values, we get:

$$h = 30 \times 3 - \frac{1}{2} \times 10 \times 3^2$$

$$h = 90 - \frac{1}{2} \times 10 \times 9$$

$$h = 90 - 45$$

$$h = 45 \text{ meters}$$

Therefore, the maximum height of the balls is 45 meters, corresponding to Option A.

Answer: Option A (45 m)

Motion

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