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 d level groun 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. Using the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\), since \(v_{0y} = 0\), we have \(h=\frac{1}{2}gt^{2}\), where \(h\) is the height of the cliff. The time of flight \(t\) for both balls is the same because they are dropped from the same height \(h\), and \(t=\sqrt{\frac{2h}{g}}\).

Step2: Analyze horizontal motion

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

Step3: Find the relationship between distances

Since \(d = 10t\) and \(d_{2}=20t\), we can express \(d_{2}\) in terms of \(d\). Divide the equation for \(d_{2}\) by the equation for \(d\): \(\frac{d_{2}}{d}=\frac{20t}{10t}=2\). So \(d_{2} = 2d\).

Answer:

\(2d\)