Calculating Upward Velocity of a Tennis Ball

This problem involves the motion of a tennis ball under constant acceleration due to gravity. We are given the maximum height reached by the ball when thrown vertically upward and need to find its initial upward velocity.

When an object is thrown vertically upward, its speed decreases due to the downward acceleration of gravity. At the maximum height, the ball momentarily stops before falling back down. This means the final velocity at the maximum height is zero.

We can use a standard kinematic equation that relates initial velocity (${u}$), final velocity (${v}$), acceleration (${a}$), and displacement (${s}$). The relevant equation is:

$$v^2 = u^2 + 2as$$

In this scenario:

• The final velocity at maximum height, ${v = 0}$ m/s.
• The displacement (maximum height), ${s = 20}$ m.
• The acceleration is due to gravity, acting downwards. Taking the upward direction as positive, the acceleration ${a = -g}$. We will use the approximate value ${g \approx 10 \, \text{m/s}^2}$ for simplicity, which is common in such problems unless specified otherwise. So, ${a = -10 \, \text{m/s}^2}$.
• The initial upward velocity, ${u}$, is what we need to find.

Now, substitute these values into the equation:

$$0^2 = u^2 + 2(-10)(20)$$

Simplify the equation:

$$0 = u^2 - 400$$

Rearrange the equation to solve for ${u^2}$:

$$u^2 = 400$$

Take the square root of both sides to find ${u}$:

$$u = \sqrt{400}$$

$$u = 20 \, \text{m/s}$$

The initial upward velocity of the tennis ball is approximately 20 m/s.

Let's compare this result with the given options:

Our calculated initial upward velocity of 20 m/s matches Option 4.

Understanding Vertical Motion and Maximum Height

When an object is projected vertically upwards, its velocity vector is initially positive (upward). The acceleration due to gravity is always negative (downward). This negative acceleration causes the magnitude of the upward velocity to decrease over time.

• The velocity becomes zero precisely at the maximum height. This is the turning point where the direction of motion changes from upward to downward.
• Above the initial point, the displacement is positive. At the maximum height, the displacement from the starting point is the maximum height value.
• The time taken to reach the maximum height is equal to the time taken to fall back to the starting point (assuming no air resistance).

Kinematic Equations for Constant Acceleration

The motion described here is motion under constant acceleration (${a = -g}$). The main kinematic equations are:

• ${v = u + at}$
• ${s = ut + \frac{1}{2}at^2}$
• ${v^2 = u^2 + 2as}$
• ${s = \frac{(u+v)}{2}t}$

We chose the third equation, ${v^2 = u^2 + 2as}$, because it directly relates the velocities (${u}$ and ${v}$), acceleration (${a}$), and displacement (${s}$), without requiring the time (${t}$).

Revision Table: Key Concepts

Additional Information: Effect of Gravity on Vertical Throw

The value of the acceleration due to gravity (${g}$) is approximately 9.8 m/s² on the surface of the Earth. However, for many introductory physics problems, especially multiple-choice questions with spread-out options, using ${g = 10}$ m/s² simplifies calculations and often leads to one of the provided options. If we had used ${g = 9.8}$ m/s² in our calculation:

$$0^2 = u^2 + 2(-9.8)(20)$$

$$0 = u^2 - 392$$

$$u^2 = 392$$

$$u = \sqrt{392} \approx 19.80 \, \text{m/s}$$

This value, 19.80 m/s, is very close to 20 m/s, confirming that 20 m/s is the correct approximate initial velocity based on the given options and the maximum height of 20 m.