Find time of flight of projectile launched from a cliff height h

Question

A projectile is launched horizontally with speed $u$ from a cliff of height $h$ . Find (a) the time of flight, and (b) the horizontal range.

Solution — Step by Step

Let the launch point be the origin. Take x-axis as horizontal (direction of launch) and y-axis as vertical (downward positive for simplicity, since the projectile falls down).

Initial conditions:

Horizontal: $u_x = u$ , $a_x = 0$ (no horizontal acceleration)
Vertical: $u_y = 0$ (launched horizontally, so no initial vertical velocity), $a_y = g = 10$ m/s² (downward)

The projectile falls a vertical distance $h$ to reach the ground.

Using $y = u_y t + \frac{1}{2}g t^2$ :

h = 0 \cdot t + \frac{1}{2}g t^2

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

Solving for $t$ :

t = \sqrt{\frac{2h}{g}}

This is the time of flight. It depends only on the height $h$ and gravity $g$ — not on the horizontal speed $u$ .

Horizontal motion is uniform (no acceleration):

R = u_x \cdot t = u \cdot \sqrt{\frac{2h}{g}}

\boxed{R = u\sqrt{\frac{2h}{g}}}

At time $t = \sqrt{2h/g}$ :

Horizontal component: $v_x = u$ (unchanged)

Vertical component: $v_y = g t = g\sqrt{2h/g} = \sqrt{2gh}$

Resultant speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + 2gh}$

This is equivalent to the result from energy conservation.

Why This Works

The key insight is independence of horizontal and vertical motions. The projectile’s horizontal motion doesn’t know anything is happening vertically, and vice versa. Horizontal: uniform motion (constant $u$ ). Vertical: free fall from rest.

The time of flight is determined entirely by the vertical drop $h$ . A heavy ball and a ping-pong ball launched horizontally from the same cliff height at the same speed hit the ground at the same time — gravity accelerates both identically (ignoring air resistance).

For a cliff problem (horizontal launch only), the initial vertical velocity is always zero. This is the key setup. If the problem says “launched at an angle,” the initial vertical velocity is $u\sin\theta$ — different scenario.

Alternative Method

Using energy conservation for the final speed:

\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh

v = \sqrt{u^2 + 2gh}

This gives the speed at impact without needing to find components separately. Note it’s the same as what we derived above.

Common Mistake

Students often think “a faster horizontal throw will take longer to fall” — that horizontal speed affects time of flight. It does not. Time of flight depends only on vertical drop: $t = \sqrt{2h/g}$ . A bullet fired horizontally and a bullet dropped vertically from the same height hit the ground at the same time (in the absence of air resistance). This is the Galileo horizontal throw experiment.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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