A cannon on a hill H = 120 m above a lake fires a projectile…

A cannon on a hill $H = 120\ \mathrm{m}$ above a lake fires a projectile horizontally toward the lake with speed $v_0 = 320\ \mathrm{m/s}$. Find where the projectile hits the water, its speed on impact, and the angle at which it strikes the surface.

Explanation of Method

A projectile fired horizontally from a hill.

Step 1: Time in the air

Horizontally the projectile keeps $v_0$; vertically it falls freely from rest: \[H = \tfrac12 g t^2 \;\Longrightarrow\; t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{240}{9.81}} \approx 4.95\ \mathrm{s}.\]

Step 2: Where it lands

\[ x = v_0 t = 320 \cdot 4.946. \]

\[x \approx 1583\ \mathrm{m} \ \text{from the foot of the hill}\]

Step 3: Velocity on impact

\[ v_x = 320\ \mathrm{m/s}, \]

\[ v_y = g t = \sqrt{2gH} \approx 48.5\ \mathrm{m/s}\ \text{(down)}, \]

\[ v = \sqrt{320^2 + 48.5^2}. \]

\[v \approx 324\ \mathrm{m/s}\]

Step 4: Angle with the water surface

\[ \tan\alpha = \frac{v_y}{v_x} = \frac{48.5}{320} \approx 0.152. \]

\[\alpha \approx 8.6^\circ \ \text{below the horizontal}\]

Check

$ v^2 = v_0^2 + 2gH = 102\,400 + 2354 = 104\,754 \Rightarrow v \approx 323.7\ \mathrm{m/s}\ \checkmark $

$H = 120\ \mathrm{m}$ $v_0 = 320\ \mathrm{m/s}$ $x \approx 1583\ \mathrm{m}$ $\text{impact}$ $v_x = 320$ $48.5$ $8.6^\circ$ $v \approx 324\ \mathrm{m/s}$ $\text{velocity at impact (to scale)}$

Path of the projectile (vertical scale stretched about 5× so the drop is visible; in reality the path is much flatter). It keeps its 320 m/s horizontally and gains 48.5 m/s downward, hitting the water at a shallow 8.6°.

Result

\[\boxed{\,x \approx 1583\ \mathrm{m} \ \text{from the foot of the hill}\,}\]

\[\boxed{\,v \approx 324\ \mathrm{m/s}\,}\]

\[\boxed{\,\alpha \approx 8.6^\circ \ \text{below the horizontal}\,}\]