Question

question 14
a small cannon is fired horizontally from the edge of a cliff that is 120 m high. the cannonball leaves the barrel with a horizontal velocity of 50 m/s.
a) how long does it take for the cannonball to hit the ground below?
b) how far from the base of the cliff will the cannonball land?
c) how far has it fallen vertically by the time it has traveled 100 m horizontally?

Explanation:

Step1: Find the time for part a

The vertical motion of the cannon - ball is a free - fall motion. The vertical displacement \(y = 120\space m\), the initial vertical velocity \(u_y=0\space m/s\), and the acceleration \(a = g= 9.8\space m/s^{2}\). Use the equation \(y=u_y t+\frac{1}{2}at^{2}\). Since \(u_y = 0\space m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\). Then \(t=\sqrt{\frac{2y}{g}}\). Substitute \(y = 120\space m\) and \(g = 9.8\space m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times120}{9.8}}\approx\sqrt{\frac{240}{9.8}}\approx4.95\space s\)

Step2: Find the horizontal distance for part b

The horizontal motion is a uniform - motion with \(v_x = 50\space m/s\) and the time \(t\) found in part a. Use the formula \(x = v_x t\). Substitute \(v_x=50\space m/s\) and \(t = 4.95\space s\) into the formula: \(x=50\times4.95 = 247.5\space m\)

Step3: Find the vertical displacement for part c

First, find the time when the horizontal displacement \(x = 100\space m\). Since \(x = v_x t\) and \(v_x=50\space m/s\), then \(t=\frac{x}{v_x}=\frac{100}{50}=2\space s\). Then use the vertical - motion formula \(y=\frac{1}{2}gt^{2}\). Substitute \(t = 2\space s\) and \(g = 9.8\space m/s^{2}\) into the formula: \(y=\frac{1}{2}\times9.8\times2^{2}=19.6\space m\)

Answer:

a) \(4.95\space s\)
b) \(247.5\space m\)
c) \(19.6\space m\)