A ball is thrown vertically upwards from the top of a building of height 25 m with a velocity of 12 m s⁻¹ - NSC Physical Sciences - Question 3 - 2022 - Paper 1

To find the time taken to reach maximum height, we can use the formula:

$$v = u + at$$

At maximum height, the final velocity (v) is 0 m/s. The initial velocity (u) is 12 m/s, and acceleration (a) is -9.8 m/s² (acting downwards). Rearranging the formula gives:

$$0 = 12 - 9.8t \ t = \frac{12}{9.8} \approx 1.22 ext{ seconds}$$

To calculate the velocity with which the ball strikes the ground, use the following kinematic equation:

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

Where:

• v = final velocity
• u = initial velocity (which is 0 m/s at the maximum height)
• a = acceleration (9.8 m/s²)
• s = total height fallen (25 m - 1.9 m = 23.1 m)

Substituting the values:

$$v^2 = 0 + 2(9.8)(23.1) \ v = \sqrt{2(9.8)(23.1)} \approx 22.67 ext{ m/s}$$

Using the formula again:

$$v = u + at$$

Here, the initial velocity (u) is the velocity just as it leaves the top of the door (which we found to be 22.67 m/s) and we can use s = height from door to ground = 1.9 m:

Rearranging for t gives:

$$t = \frac{v - u}{a} = \frac{0 - 22.67}{-9.8} \approx 2.31 ext{ seconds}$$

To draw the velocity vs. time graph:

1. The graph should start at +12 m/s (the initial upward velocity).
2. The velocity decreases linearly to 0 m/s at the maximum height (1.22 seconds).
3. After reaching maximum height, the velocity becomes negative, indicating downward motion.
4. The velocity increases negatively until it strikes the ground, reaching about -22.67 m/s just before impact.

The graph should clearly indicate:

• Peak at 1.22 seconds
• Show time intervals appropriately
• Include the initial upward velocity, time to maximum height, and final velocity upon striking the ground.