Q.A hiker stands on the edge of a cliff $490\ \text{m}$ above the ground and throws a stone horizontally with an initial speed of $15\ \text{m s}^{-1}$. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take $g = 9.8\ \text{m s}^{-2}$).

Concept: Projectile Motion — the vertical and horizontal motions are independent; time of flight depends only on vertical drop.

1. Time to reach ground — vertical motion starts from rest:

   $s_y = \frac12 g t^2$

   $490 = \frac12 (9.8) t^2$

   $t^2 = \frac{490 \times 2}{9.8} = 100$

   $t = 10\ \text{s}$

2. Vertical speed on impact:

   $v_y = g t = 9.8 \times 10 = 98\ \text{m s}^{-1}$

3. Resultant speed — horizontal speed remains constant at $15\ \text{m s}^{-1}$: …

Approach

Because the stone is thrown horizontally, its initial vertical velocity is zero — vertically it is in free fall from rest, exactly as if simply dropped. The horizontal velocity stays constant at $15\ \text{m s}^{-1}$ (no air resistance). The time to reach the ground is set entirely by the vertical drop.

Step 1 — Time of fall

Taking downward as positive with $u_y = 0$:

Step 2 — Velocity components at impact

Vertical component:

Horizontal component (unchanged):

$v_x = 15\ \text{m s}^{-1}$ …

Concept: Independence of Motions for Time, Work-Energy Theorem for Speed

Method: Two Different Principles for the Two Parts -- Galileo's Independence Principle (time), Energy Conservation (speed)

Instead of finding both the vertical and horizontal velocity components and combining them by Pythagoras, this method finds the impact speed directly from the work-energy theorem -- equating the kinetic energy gained to the gravitational potential energy lost -- without ever computing $v_y$ and $v_x$ separately.

Part (a) -- Time of fall, via independence of motions

1. Galileo's principle: the horizontal throw does not affect the vertical fall at all. A stone thrown horizontally at $15\ \text{m s}^{-1}$ takes exactly as long to fall $490\ \text{m}$ as a stone simply dropped from rest at the same height -- the horizontal motion is a completely independent, parallel story.

2. So treat it as a pure vertical drop from rest:

Part (b) -- Impact speed, via the work-energy theorem (no velocity components at all)

3. State the work-energy theorem for the whole fall, taking the ground as the reference level for potential energy: the loss in gravitational PE equals the gain in KE (air resistance neglected, so mechanical energy is conserved):

where $u=15\ \text{m s}^{-1}$ is the initial (purely horizontal) speed and $v$ is the final speed -- the mass $m$ cancels from every term.

4. Solve for $v$:

Why energy conservation is a genuine shortcut for the speed …