A player throws a ball upwards with an initial speed of 29.4m s⁻¹. (a) What is the direction of acceleration d

Exercises · 2.6

29.4\ \text{m s}^{-1}

(a) What is the direction of acceleration during the upward motion of the ball?

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

x = 0\ \text{m}

and

t = 0\ \text{s}

to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of

x

-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

(d) To what height does the ball rise and after how long does the ball return to the player's hands? (Take

g = 9.8\ \text{m s}^{-2}

and neglect air resistance).

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Concept understanding — Motion Under Gravity

Motion Under Gravity – The Intuition First

Drop a ball. It falls. Why? Because the Earth pulls it. But the key insight is not just that it falls — it's how it falls. If you drop a stone from a cliff, does it move at a constant speed? No. It gets faster as it goes down. That's the heart of motion under gravity: the pull is steady, so the acceleration is steady.

Now imagine throwing a ball straight up. It rises, slows down, stops for an instant at the top, then falls back down. The same pull that made it slow down on the way up makes it speed up on the way down. The acceleration never changes direction — it always points downward, toward the centre of the Earth.

That constant downward acceleration is called $g$ . Near the Earth's surface, its magnitude is about $9.8\ \text{m/s}^2$ (often taken as $10\ \text{m/s}^2$ in exams for quick calculation). The direction is always downward.

Important

In one-dimensional motion under gravity, the acceleration is constant and equal to $g$ downward. This is true only when air resistance is negligible and the height is small compared to the Earth's radius.

The Precise Statement

Motion under gravity (in one dimension) means an object moves vertically under the sole influence of the Earth's gravitational pull. The acceleration is:

a = -g

where the negative sign indicates downward direction (if we take upward as positive). The value of $g$ is $9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ in many problems).

Because acceleration is constant, all three equations of uniformly accelerated motion apply directly:

v = u + at

s = ut + \frac12 at^2

v^2 = u^2 + 2as

But here $a$ is replaced by $-g$ (if upward is positive) or $+g$ (if downward is positive). The choice of sign convention is yours — just be consistent.

v = u - gt

Key Features You Must Know

1. Free Fall (dropped from rest)

If you simply let go of an object ( $u = 0$ ), it falls with increasing speed. After time $t$ , its velocity is $v = gt$ (downward). The distance fallen is $s = \frac12 gt^2$ .

2. Projected Upward

If you throw a ball upward with speed $u$ , it rises until its velocity becomes zero at the highest point. That takes time $t = u/g$ . The maximum height reached is $h = u^2/(2g)$ .

3. Symmetry of the Motion

The time to go up equals the time to come back down (to the same height). The speed at which it returns to the launch point equals the initial speed $u$ (but downward). This symmetry is a direct consequence of constant acceleration.

Watch out

At the highest point, velocity is zero but acceleration is still $g$ downward. Many beginners think acceleration becomes zero at the top — it does not. The object is momentarily at rest, but gravity is still pulling it.

A Simple Example …

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Concept: Projectile Motion Under Gravity — acceleration is constant and always directed downward.

(a) The acceleration is due to gravity, which always acts vertically downward.

Direction: downward.

(b) At the highest point, the vertical velocity becomes zero. Acceleration remains $g = 9.8\ \text{m s}^{-2}$ downward.

Velocity: $0\ \text{m s}^{-1}$ ; Acceleration: $9.8\ \text{m s}^{-2}$ downward.

(c) With $x=0$ at the highest point and positive $x$ taken downward, the ball is always below the highest point during both phases of its flight (it can only equal $x=0$ at the single instant it is at the top), so its position is positive throughout:

Motion	Position $x$	Velocity $v$	Acceleration $a$
Upward (rising toward $x=0$ )	positive	negative (moving opposite to $+x$ )	positive
Downward (falling away from $x=0$ )	positive	positive (moving along $+x$ )	positive

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This problem explores the symmetry of projectile motion under constant gravity. The acceleration is always downward ( $9.8\ \text{m/s}^2$ ), velocity is zero at the top, and the ball rises $44.1\ \text{m}$ in $3\ \text{s}$ , taking another $3\ \text{s}$ to return — total flight time $6\ \text{s}$ .

The Core Idea: Motion Under Constant Gravity

When you throw a ball upward, the only force acting on it (ignoring air resistance) is gravity. This force produces a constant downward acceleration of $g = 9.8\ \text{m/s}^2$ throughout the entire motion — whether the ball is going up, has stopped at the top, or is coming back down. This is the single most important fact to internalize.

The motion is perfectly symmetric: the time to go up equals the time to come down, and the speed at any height on the way up equals the speed at the same height on the way down.

(a) Direction of acceleration during upward motion

The ball is moving upward, but gravity pulls it downward. Acceleration due to gravity is always directed toward the Earth's center — vertically downward. This never changes.

Watch out

A common mistake is to think acceleration is upward while the ball rises. It is not. Acceleration is always downward, even when velocity is upward. The ball slows down because acceleration opposes velocity.

Answer (a): The direction of acceleration is vertically downward throughout the upward motion.

(b) Velocity and acceleration at the highest point

At the highest point, the ball momentarily stops before reversing direction. Its velocity becomes zero for an instant.

But gravity doesn't take a break. The acceleration remains $g = 9.8\ \text{m/s}^2$ downward, even at that instant.

Tip

Think of it this way: if acceleration were zero at the top, the ball would hover there. Instead, it immediately begins falling — proof that acceleration is still present.

Answer (b): At the highest point, velocity is $0\ \text{m/s}$ and acceleration is $9.8\ \text{m/s}^2$ downward.

(c) Sign convention for position, velocity, and acceleration

We set:

$x = 0$ at the highest point
Positive $x$ -axis = vertically downward
$t = 0$ at the highest point

The ball starts from the player's hand, rises to $x=0$ (the highest point), then falls back down. Since downward is positive and the highest point is the origin, the ball can never be above $x=0$ — it is at $x=0$ only at the single instant it is at the top, and is at positive $x$ (below the top) during both the rise (before reaching the top) and the fall (after leaving the top).

Phase	Position $x$	Velocity $v$	Acceleration $a$
Upward (rising toward the top)	Positive (below the top)	Negative (moving opposite to $+x$ , i.e. upward)	Positive (downward = $+x$ direction)
Downward (falling away from the top)	Positive (below the top)	Positive (moving in $+x$ direction, i.e. downward)	Positive

Check: At the highest point itself, $x=0$ , $v=0$ , $a=+g$ . …

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Concept: Reading an Entire Multi-Part Question off One v-t Graph

Method: The Single v-t Line (slope, zero-crossing, symmetry, and area together)

All four parts of this question — direction of acceleration, velocity/acceleration at the peak, sign behaviour, and the numerical height/time — can be answered from one single straight-line graph: $v(t) = u - gt$ with $u=29.4\ \text{m s}^{-1}$ (upward positive), instead of applying a separate formula to each part in turn.

Steps

Draw the graph. $v(t) = 29.4 - 9.8t$ is one continuous straight line for the entire flight (up and down) — it never bends, because gravity never switches off or changes magnitude.
(a) Direction of acceleration — read it from the line's slope. The slope is $-9.8\ \text{m s}^{-2}$ everywhere along the line, including the portion where $v>0$ (ball rising). A constant negative slope on this graph means acceleration is constant and downward throughout — even while the ball moves upward.
(b) Velocity and acceleration at the highest point — read it from where the line crosses the time axis. Setting $v=0$ : $0 = 29.4-9.8t \Rightarrow t=3\ \text{s}$ . At that instant, $v=0$ by definition of the crossing, but the line's slope there is still $-9.8\ \text{m s}^{-2}$ (a straight line has the same slope everywhere) — so acceleration is $9.8\ \text{m s}^{-2}$ downward even at the very instant $v=0$ .
(c) Sign convention with origin at the top. Re-plot with $x=0$ , $t=0$ at the peak and downward positive: since the ball is below the peak at every instant except the peak itself, $x>0$ throughout both halves of the flight. Velocity is negative while rising toward the peak (opposing the new $+x$ = downward direction) and positive while falling away from it; acceleration stays positive (downward) throughout, consistent with Step 2's constant-slope observation.
(d) Maximum height — read it from the area under the graph, from $t=0$ to $t=3\ \text{s}$ (the part where $v>0$ ). This region is a triangle with base $3\ \text{s}$ and height $29.4\ \text{m s}^{-1}$ :

$h = \frac{1}{2}(3)(29.4) = 44.1\ \text{m}$ …

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Q.A player throws a ball upwards with an initial speed of 29.4 m s−129.4\ \text{m s}^{-1}29.4 m s−1.

The Core Idea: Motion Under Constant Gravity

(a) Direction of acceleration during upward motion

(b) Velocity and acceleration at the highest point

(c) Sign convention for position, velocity, and acceleration

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Q.A player throws a ball upwards with an initial speed of $29.4\ \text{m s}^{-1}$ .

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