A ball is thrown horizontally from 80m height at 10 m/s — where does it land

Question

A ball is thrown horizontally from a height of 80 m with an initial speed of 10 m/s. Find: (a) the time of flight, and (b) the horizontal range (how far from the base of the cliff does it land). (Take $g = 10$ m/s²)

Solution — Step by Step

This is a horizontal projectile problem. A key feature of projectile motion is that horizontal and vertical motions are completely independent — they don’t affect each other.

Horizontal: Constant velocity (no horizontal force, ignoring air resistance). $v_x = 10$ m/s (constant throughout flight).

Vertical: Free fall with acceleration $g = 10$ m/s² downward. Initial vertical velocity = 0 (thrown horizontally, not upward or downward).

The ball falls 80 m vertically (from 80 m height to ground at 0 m). Using the equation of motion:

h = u_y t + \frac{1}{2}g t^2

With $h = 80$ m (downward), $u_y = 0$ , $g = 10$ m/s²:

80 = 0 + \frac{1}{2}(10)t^2 = 5t^2

t^2 = \frac{80}{5} = 16

t = 4 \text{ s}

The ball is in flight for 4 seconds.

During these 4 seconds, the ball moves horizontally at a constant 10 m/s:

R = v_x \times t = 10 \times 4 = 40 \text{ m}

The ball lands 40 metres from the base of the cliff.

At landing:

Horizontal velocity: $v_x = 10$ m/s (unchanged)
Vertical velocity: $v_y = u_y + gt = 0 + 10 \times 4 = 40$ m/s (downward)

Resultant speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{100 + 1600} = \sqrt{1700} \approx 41.2$ m/s

Angle below horizontal: $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{40}{10}\right) = \tan^{-1}(4) \approx 76°$

Time of flight: $t = 4$ s
Horizontal range: $R = 40$ m
(Bonus) Landing speed: ~41.2 m/s at ~76° below horizontal

Why This Works

The independence of horizontal and vertical motion is the core principle of projectile motion. Gravity only acts vertically — it accelerates the ball downward but has no horizontal component (in ideal projectile motion without air resistance). So:

Horizontal: $x = v_x t$ (uniform motion — no force, no acceleration)
Vertical: $y = \frac{1}{2}gt^2$ (uniform acceleration from rest — gravity only)

These two equations are solved independently and combined at the end to find where the ball is at any time $t$ .

Alternative Method — Energy Conservation for Landing Speed

Using energy conservation to find landing speed (skips finding vertical velocity separately):

At launch: KE = $\frac{1}{2}mv_x^2 = \frac{1}{2}m(100)$ . PE = $mgh = m(10)(80) = 800m$ .

At landing: All PE converted to KE: Total KE = $\frac{1}{2}mv_x^2 + mgh = 50m + 800m = 850m$

$\frac{1}{2}mv^2 = 850m \Rightarrow v = \sqrt{1700} \approx 41.2$ m/s ✓

Same answer through energy — a useful check.

Common Mistake

The most common error is using the initial velocity (10 m/s) in the vertical equation or forgetting that the initial vertical velocity is zero. For a horizontally thrown projectile, $u_y = 0$ — there is no vertical component to the launch velocity. The 10 m/s is entirely horizontal.

Another mistake: computing $R = v_x \times t$ with the wrong time. Always find $t$ from the vertical equation first, then use it in the horizontal equation. Do not guess the time.

For any horizontal projectile:

Time of flight depends ONLY on height and $g$ : $t = \sqrt{2h/g}$
Range depends on $v_x$ and $t$ : $R = v_x \sqrt{2h/g}$

Memorising these two combined formulas can save time in MCQs: $t = \sqrt{2 \times 80/10} = \sqrt{16} = 4$ s and $R = 10 \times 4 = 40$ m.

Question

Solution — Step by Step

Why This Works

Alternative Method — Energy Conservation for Landing Speed

Common Mistake

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