Let's move on to Question 7.

Question 7: Vertical Motion Under Gravity

A tennis ball is thrown vertically upwards with an initial velocity of 6 m/s from a point that is 1.5 m above the ground.

7.1 Calculate the maximum height that the tennis ball reaches above the ground.

To calculate the maximum height, we use the kinematic equation for vertical motion under constant acceleration (gravity): $v_f^2 = v_i^2 + 2a \Delta y$ Where:

• $v_f = 0 \, \text{m/s}$ (final velocity at the highest point),
• $v_i = 6 \, \text{m/s}$ (initial velocity),
• $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity, negative because it's acting downward),
• $\Delta y$ is the displacement (height reached above the initial position).

Rearranging the equation to solve for $\Delta y$: $0 = (6)^2 + 2(-9.8) \Delta y$ $0 = 36 - 19.6 \Delta y$ $19.6 \Delta y = 36$ $\Delta y = \frac{36}{19.6} \approx 1.84 \, \text{m}$

The ball reaches 1.84 m above its starting point.

Since the ball started from a height of 1.5 m above the ground, the total maximum height ( h