A ball is thrown vertically upwards - Scottish Highers Physics - Question 3 - 2017

To calculate the distance the ball falls, we can use the following kinematic equation:

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

Where:

• $v$ = final velocity just before hitting the ground (-7 m/s)
• $u$ = initial velocity (0 m/s at maximum height)
• $a$ = acceleration due to gravity (-9.8 m/s²)
• $s$ = distance fallen

Plugging in the values, we have:

$(-7)^2 = 0^2 + 2(-9.8)s$

This simplifies to:

$49 = -19.6s$

Thus,

$s = \frac{-49}{-19.6} \approx 2.5 ext{ m}$

Hence, the distance the ball falls from its maximum height to the ground is approximately 2.5 meters.

To indicate a greater initial vertical velocity, we would add another line to the graph. This new line would start from the same initial height but would reflect a steeper descent as it would be thrown upwards with more force. The line would remain positive for a shorter time before turning downwards, indicating a more pronounced falling phase. The new graph would not require new numerical values on the axes, but should visibly illustrate the greater initial velocity through a steeper slope above the original line.