Question

1. a cannon is fired horizontally from the top of a 20m cliff into the ocean. the initial velocity of the cannon was 100m/s. how far from the cliff does the cannon - ball hit the water?

Explanation:

Step1: Analyze vertical - motion

The ball is in free - fall vertically. The vertical displacement $y = 30m$, the initial vertical velocity $u_y=0m/s$, and the acceleration due to gravity $g = 9.8m/s^{2}$. Using the equation $y=u_y t+\frac{1}{2}gt^{2}$, since $u_y = 0m/s$, we have $y=\frac{1}{2}gt^{2}$.

Step2: Solve for time $t$

From $y=\frac{1}{2}gt^{2}$, we can re - arrange to get $t=\sqrt{\frac{2y}{g}}$. Substituting $y = 30m$ and $g = 9.8m/s^{2}$, we have $t=\sqrt{\frac{2\times30}{9.8}}\approx\sqrt{6.12}\approx2.47s$.

Step3: Analyze horizontal - motion

The horizontal velocity $v_x = 10m/s$ (constant because there is no horizontal acceleration). Using the formula $x = v_x t$.

Step4: Calculate horizontal distance

Substitute $v_x = 10m/s$ and $t\approx2.47s$ into $x = v_x t$. So $x=10\times2.47 = 24.7m$.

Answer:

$24.7m$