Physics 9702/22 — May/June 2011

Background Concept

Projectile motion can be split into two independent perpendicular motions:

• Horizontal: constant velocity (no horizontal acceleration when air resistance is negligible).
• Vertical: constant acceleration $g$ downward.

For vertical free fall starting with no vertical component of velocity, the displacement $s$ after time $t$ is

$$s = ut + \frac{1}{2}at^2$$

With $u=0$ and $a=g$, this becomes $s = \tfrac{1}{2}gt^2$.

Understanding the Question

The stone is thrown horizontally, so initially its vertical velocity is zero. It falls a vertical distance of $15\ \text{m}$ under gravity, with air resistance negligible. You must find the time taken to fall that $15\ \text{m}$.

Approach

Use vertical motion only:

1. Set $u_y = 0$.
2. Use $s = \tfrac{1}{2}gt^2$.
3. Rearrange for $t$.

Step-by-Step Reasoning

Take downward as positive for convenience.

Given:

• $s = 15\ \text{m}$
• $u_y = 0$
• $a = g = 9.81\ \text{m s}^{-2}$

Use

$$15 = \frac{1}{2}(9.81)t^2$$

So

$$t^2 = \frac{30}{9.81} \approx 3.06$$

and

$$t = \sqrt{3.06} \approx 1.75\ \text{s}$$

(The horizontal speed does not affect the fall time because horizontal and vertical motions are independent.)

Key Takeaways

• Horizontal projection means $u_y = 0$.
• Time to fall depends only on vertical distance and $g$, not on horizontal speed.

Common Mistakes

• Using $u = 20\ \text{m s}^{-1}$ in the vertical equation (that is the horizontal component).
• Using $s = gt^2$ (missing the factor $\tfrac{1}{2}$).

Things to Be Careful About

• Keep the vertical distance as $15\ \text{m}$ (already in SI units).
• Use a consistent value of $g$ (commonly $9.81\ \text{m s}^{-2}$ or $9.8\ \text{m s}^{-2}$ depending on the paper).

Background Concept

In projectile motion with negligible air resistance:

• Horizontal acceleration $a_x = 0$ so horizontal velocity stays constant.
• Vertical acceleration $a_y = g$ so vertical velocity changes uniformly.

The actual velocity at any instant is a vector with components $v_x$ and $v_y$. The magnitude (speed) is found by Pythagoras because the components are perpendicular:

$$v = \sqrt{v_x^2 + v_y^2}$$

Understanding the Question

After falling $15\ \text{m}$, the stone has gained a downward vertical velocity due to gravity but keeps the same horizontal velocity of $20\ \text{m s}^{-1}$. You must calculate the magnitude of the resultant velocity at that point.

Approach

1. Keep $v_x = 20\ \text{m s}^{-1}$ (no horizontal acceleration).
2. Find $v_y$ using $v_y = u_y + gt$ (or $v_y^2 = u_y^2 + 2gs$).
3. Combine $v_x$ and $v_y$ using Pythagoras to get the resultant speed.

Step-by-Step Reasoning

Horizontal component:

$$v_x = 20\ \text{m s}^{-1}$$

Vertical component: initial vertical velocity $u_y = 0$. Using the time from part (i), $t = 1.75\ \text{s}$,

$$v_y = u_y + gt = 0 + (9.81)(1.75) \approx 17.2\ \text{m s}^{-1}$$

Now form the velocity triangle:

Magnitude:

$$v = \sqrt{(20)^2 + (17.2)^2}$$ $$v = \sqrt{400 + 296} \approx \sqrt{696} \approx 26.4\ \text{m s}^{-1}$$

Key Takeaways

• $v_x$ stays constant without air resistance.
• $v_y$ increases from $0$ due to gravity.
• Resultant speed comes from combining perpendicular components.

Common Mistakes

• Adding components directly: $v = 20 + 17.2$ (wrong because velocities are perpendicular vectors).
• Using $20\ \text{m s}^{-1}$ as the initial vertical velocity.
• Forgetting that the question asks for magnitude (so no direction is required in the final answer).

Things to Be Careful About

• Use consistent significant figures (typically 2–3 s.f.).
• If you use the alternative method $v_y^2 = 2gs$, you should still combine with $v_x$ using Pythagoras.
• Keep units throughout: $\text{m s}^{-1}$ for velocities.

Background Concept

• Displacement is a vector: it describes the change in position from start to finish. It is the straight-line vector joining the initial and final positions.
• Distance travelled is a scalar: it is the total length of the path actually taken.

For curved motion, distance and the magnitude of displacement are not the same.

Understanding the Question

The stone follows a curved (parabolic) path from the top of the cliff to the ground. The question asks you to describe how:

• the displacement (start-to-finish vector), and
• the distance travelled (length along the curve)
  are different for this motion.

Approach

State:

1. what displacement means here (straight line between initial and final points + direction),
2. what distance means here (length along the trajectory),
3. compare them (distance is larger than the magnitude of displacement for a non-straight path).

Step-by-Step Reasoning

• The displacement is drawn from the throw point at the top of the cliff directly to the landing point on the ground: it does not follow the curve. Because it is a vector, it has a direction (downwards and horizontally away from the cliff).
• The distance travelled is measured along the projectile’s parabolic trajectory, which is longer than the straight-line separation.

A sketch can help:

Key Takeaways

• Displacement: straight-line vector from start to end.
• Distance: total path length (scalar).
• For curved motion, $\text{distance} > |\text{displacement}|$.

Common Mistakes

• Saying displacement is “how far it moves” (that describes distance, not displacement).
• Forgetting to mention that displacement is a vector (direction matters).
• Claiming distance and displacement are the same because the object “moves from A to B” (only true if the path is a straight line).

Things to Be Careful About

• Displacement depends only on the initial and final positions, not on the route.
• The question asks for a description, not a calculation, so clear definitions earn the marks.