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Constant acceleration refers to a scenario where the rate of change in velocity is the same throughout the time period under consideration. In the context of this exercise, the car starts from rest with a constant acceleration of \(2 \mathrm{\ m/s^2}\). This means every second, the velocity of the car increases by 2 meters per second.
Initially, the car's velocity is zero, since it starts from rest. Over time, as it continues to accelerate at this constant rate, its speed grows linearly. For example, after 5 seconds, the car achieves a speed of \(10 \mathrm{\ m/s}\), calculated using the formula:

• \( v = u + at \)

Here, \(u\) is the initial velocity (0 in this case), \(a\) is the acceleration, and \(t\) is time. Understanding constant acceleration is crucial for solving motion problems, especially when dissecting projectile motion.

Vertical velocity is a component of an object's velocity in the vertical direction. When the ball is dropped from the car, it initially has no vertical speed because it isn't thrown downward; it begins dropping from rest.
As soon as the ball is released, the force of gravity acts on it, imparting acceleration downwards at \(10 \mathrm{\ m/s^2}\). As a result, its vertical velocity increases over time.
To find this velocity after \(0.5\) seconds, we use:

• \( v_y = u_y + g \times t \)

Where \(u_y\) is the initial vertical velocity (0), \(g\) is the acceleration due to gravity, and \(t\) is time. This gives:
\(v_y = 0 + 10 \times 0.5 = 5 \mathrm{\ m/s}\).
This shows how gravity affects a projectile's vertical velocity over time.

Horizontal velocity is the velocity of an object moving parallel to the horizontal plane. Unlike vertical velocity, horizontal velocity remains constant if there's no external force acting on it, such as air resistance.
In this exercise, the ball inherits the car's horizontal speed at the moment it's released. This speed is maintained since horizontal motion is unaffected by gravity.
The horizontal velocity, \(v_x\), is thus the same as the car's speed at the time of release, which is \(10 \mathrm{\ m/s}\).
Understanding horizontal velocity is key to analyzing projectile motion, as it allows us to isolate and evaluate the distinct forces acting on the object in motion.

The resultant velocity is the combined velocity of an object considering both its horizontal and vertical components. It gives us the actual speed and direction of an object in projectile motion.
To determine the resultant velocity of the ball, we combine its horizontal velocity, \(v_x = 10 \mathrm{\ m/s}\), and its vertical velocity, \(v_y = 5 \mathrm{\ m/s}\), at \(0.5\) seconds after release.
The formula to calculate the resultant velocity is:

• \( v = \sqrt{v_x^2 + v_y^2} \)

By substituting the known values:
\( v = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \mathrm{\ m/s} \).
This resultant velocity helps us understand the ball's actual trajectory and how both gravity and initial conditions affect its path.