Question

1. the diagram below represents a ball projected from a cliff at a speed of 10. meters per second. the ball travels the path shown and lands at time t and distance d from the base of the cliff. v = 10. m/s time = t seconds level ground a second identical ball is projected horizontally from the cliff at 20. meters per second. determine the distance the second ball lands from the base of the cliff in terms of d.

Explanation:

Step1: Analyze vertical motion

The vertical - motion of both balls is free - fall. The initial vertical velocity \(v_{0y}=0\ m/s\) for both balls, and the vertical displacement \(y = h\) (height of the cliff). Using the equation \(y=v_{0y}t+\frac{1}{2}gt^{2}\), since \(v_{0y} = 0\), we have \(y=\frac{1}{2}gt^{2}\), and the time of flight \(t=\sqrt{\frac{2y}{g}}\). Since the height of the cliff \(y\) is the same for both balls, the time of flight \(t\) is the same for both balls.

Step2: Analyze horizontal motion

For the first ball, the horizontal velocity is \(v_{1x}=10\ m/s\), and the horizontal displacement \(d = v_{1x}t=10t\). For the second ball, the horizontal velocity is \(v_{2x}=20\ m/s\), and the horizontal displacement \(d_{2}=v_{2x}t = 20t\).

Step3: Find the relationship between \(d\) and \(d_{2}\)

We know \(d = 10t\) and \(d_{2}=20t\). Dividing the second equation by the first equation: \(\frac{d_{2}}{d}=\frac{20t}{10t}=2\), so \(d_{2}=2d\).

Answer:

\(2d\)