Q.A ball is thrown vertically upwards with a velocity of $20\ \text{m s}^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0\ \text{m}$ from the ground.

Concept: Projectile motion under gravity — vertical motion only, with constant acceleration $g = 10\ \text{m s}^{-2}$ downward.

1. Maximum height from the throw point: At the highest point, final velocity $v = 0$. Using $v^2 = u^2 - 2gh$: $0 = (20)^2 - 2(10)h \implies h = \frac{400}{20} = 20\ \text{m}$. So the ball rises $20\ \text{m}$ above the building top.
2. Time to hit the ground: Total displacement from throw point to ground is $-25\ \text{m}$ (downward). Using $s = ut - \frac12 gt^2$: …

The key idea here is that the ball moves under constant acceleration due to gravity ($g = 10\ \text{m/s}^2$ downward). We choose a sign convention: upward is positive, downward is negative. So acceleration is $a = -g = -10\ \text{m/s}^2$.

We treat the motion in two natural phases: the upward journey until the ball stops, and then the downward journey from that highest point to the ground. But we can also solve the whole motion in one go using the equations of motion — that's what we'll do for part (b).

Figure 2.6 shows this exact setup: the building drawn as a solid bar 25 m tall, with the throw point at the top labelled A and a dashed curved path rising to the highest point B, where $(y - y_0)$ marks the extra height gained above A. The origin $y = 0$ is at the ground, so the ball's full height above the ground is the longer bracket $y$; point C marks the foot of the building directly below A. The downward arrow labelled $a = -10\ \text{m/s}^2$ is a reminder that gravity acts throughout, including while the ball is still rising.

1. Part (a): How high will the ball rise?

At the highest point, the ball's velocity becomes zero for an instant. We know:

• Initial velocity $u = +20\ \text{m/s}$
• Final velocity $v = 0$
• Acceleration $a = -10\ \text{m/s}^2$
• Displacement $s = h$ (height above the throw point)

Use the equation $v^2 = u^2 + 2as$:

So the ball rises 20 metres above the point of throw.

2. Part (b): How long before the ball hits the ground? …

Concept: Splitting Vertical Motion into Physical Phases

Method: Two-Phase Decomposition (up, then down) instead of one combined equation

Rather than writing a single displacement equation covering the ball's entire trip from the throw point to the ground (which needs the quadratic formula), this method treats the "rise" and the "fall" as two separate, simpler free-fall problems, using the ball's momentary rest at the top as the natural break point.

Take upward as positive, $u = 20\ \text{m s}^{-1}$, $g = 10\ \text{m s}^{-2}$, building height $H = 25.0\ \text{m}$.

Phase 1 — Rise to the highest point

1. At the top, velocity is momentarily zero. Time to get there, from $v = u - gt_1$ with $v=0$:

2. Height risen above the throw point, from $v^2 = u^2 - 2gh$ with $v=0$:

Phase 2 — Fall from the highest point to the ground

3. The highest point is $h = 20\ \text{m}$ above the throw point, which is itself $H = 25\ \text{m}$ above the ground — so the ball must now fall a fresh, independent distance:

4. This is now a simple drop from rest (velocity is exactly zero at the start of this phase — that's the whole point of splitting it here). Using $H_{\text{fall}} = \tfrac{1}{2}g t_2^2$: …