Understanding the Problem: Vertical Projectile Motion

The question asks for the magnitude of the displacement of a ball thrown vertically upwards after a specific time. We are given the initial velocity, the time elapsed, and the acceleration due to gravity. This is a classic problem involving motion under constant acceleration, specifically, motion in one dimension (vertical).

Key Concepts for Vertical Throw

• When an object is thrown vertically upwards, its velocity decreases as it rises due to the downward force of gravity.
• At the highest point, the instantaneous velocity of the object becomes zero.
• As it falls back down, its velocity increases in the downward direction.
• The acceleration due to gravity ($g$) acts downwards throughout the motion.
• We can use kinematic equations (equations of motion) to describe the position and velocity of the object at any time.

Identifying Given Information

From the problem statement, we have:

• Initial velocity, $u = 40$ m/s (upwards)
• Time elapsed, $t = 6$ s
• Acceleration due to gravity, $g = 10$ m/s² (downwards)

Since we are considering upward as the positive direction (because the initial velocity is upwards), the acceleration due to gravity acting downwards will be negative.

Therefore, acceleration, $a = -g = -10$ m/s².

Choosing the Correct Kinematic Equation

We need to find the displacement ($s$) after time ($t$). The kinematic equation that relates initial velocity ($u$), time ($t$), acceleration ($a$), and displacement ($s$) is:

$$s = ut + \frac{1}{2}at^2$$

Calculating the Displacement

Now, let's substitute the given values into the equation:

$$s = (40 \text{ m/s})(6 \text{ s}) + \frac{1}{2}(-10 \text{ m/s}^2)(6 \text{ s})^2$$

First, calculate the terms:

$$s = 240 \text{ m} + \frac{1}{2}(-10 \text{ m/s}^2)(36 \text{ s}^2)$$

$$s = 240 \text{ m} + (-5 \text{ m/s}^2)(36 \text{ s}^2)$$

$$s = 240 \text{ m} - 180 \text{ m}$$

$$s = 60 \text{ m}$$

The displacement calculated is $+60$ m. The positive sign indicates that the final position of the ball after 6 seconds is 60 meters above the starting point.

Understanding Displacement vs. Distance

Displacement is a vector quantity representing the change in position from the starting point to the ending point. Distance is a scalar quantity representing the total path length traveled. In this case, the ball goes up, reaches its maximum height, and then falls back down. The displacement is simply the vertical distance from the initial position to the final position after 6 seconds.

Determining the Magnitude of Displacement

The magnitude of the displacement is the absolute value of the displacement. In this case, the displacement is $60$ m. The magnitude is $|60 \text{ m}| = 60$ m.

Summary of Steps

1. Identify the known variables: initial velocity ($u$), time ($t$), and acceleration ($a = -g$).
2. Choose the appropriate kinematic equation: $s = ut + \frac{1}{2}at^2$.
3. Substitute the values into the equation.
4. Calculate the displacement.
5. Determine the magnitude of the displacement.

Using $s = ut + \frac{1}{2}at^2$:

$$s = (40)(6) + \frac{1}{2}(-10)(6)^2$$

$$s = 240 - 5(36)$$

$$s = 240 - 180$$

$$s = 60 \text{ m}$$

The magnitude of the displacement is 60 m.

Revision Table: Vertical Motion Concepts

Additional Information: Maximum Height and Time of Flight

Let's explore some additional aspects of this vertical throw problem that are related.

Maximum Height

To find the maximum height, we know that the velocity at the peak is zero ($v=0$). We can use the equation $v^2 = u^2 + 2as_{max}$.

$$0^2 = (40 \text{ m/s})^2 + 2(-10 \text{ m/s}^2)s_{max}$$

$$0 = 1600 \text{ m}^2/\text{s}^2 - 20 \text{ m/s}^2 \cdot s_{max}$$

$$20 s_{max} = 1600$$

$$s_{max} = \frac{1600}{20} = 80 \text{ m}$$

The maximum height reached by the ball is 80 m above the starting point.

Time to Reach Maximum Height

We can find the time it takes to reach the maximum height using $v = u + at_{top}$. At the top, $v=0$.

$$0 = 40 \text{ m/s} + (-10 \text{ m/s}^2)t_{top}$$

$$10 t_{top} = 40$$

$$t_{top} = 4 \text{ s}$$

It takes 4 seconds for the ball to reach its maximum height.

Analyzing Displacement at 6 Seconds

Since the time to reach maximum height is 4 seconds, at $t=6$ seconds, the ball has already reached its peak (at $t=4$ s) and is falling back down for another $6 - 4 = 2$ seconds.

After reaching 80 m height at $t=4$ s, the ball falls for 2 seconds. We can calculate the distance fallen in these 2 seconds using $s_{fall} = u_{fall}t' + \frac{1}{2}at'^2$, where $u_{fall}=0$ (velocity at the top), $t'=2$ s, and $a=10$ m/s² (taking downward as positive for this part, or sticking to upward as positive, $u_{fall}=0$ and $a=-10$, $t'=2$ s).

Using $u=0$ and $a=-10$ m/s² for the 2-second fall starting from the top (initial position for this segment is the peak at 80m):

Displacement from the peak: $s' = (0 \text{ m/s})(2 \text{ s}) + \frac{1}{2}(-10 \text{ m/s}^2)(2 \text{ s})^2$

$$s' = 0 + \frac{1}{2}(-10)(4) = -20 \text{ m}$$

The negative sign means the displacement from the peak is downwards by 20 m. The peak was at 80 m from the start. So, the position after 6 seconds is $80 \text{ m} - 20 \text{ m} = 60$ m above the start. This confirms the displacement calculation using the total time.

The magnitude of the displacement remains 60 m.