A speed boat is being chased down a river by second boat that is faster. Just as the second boat pulls up to first boat, they reach the edge of a 5.0 m cliff, first boat at 15m/s and second at 26m/s. How far apart will they be when they land below?
Let us assume that when both boats reach the edge of the cliff, their velocities are horizontal, since otherwise is not specified in the problem. This means, it will take them the same time to land 5 below the cliff. This time is determined by the equation
`h=(g*t^2)/2` , where h is the height h = 5m, g is the acceleration of the free fall g = 9.8 m/s^2, and t is the time. From here,
`t = sqrt((2h)/g) =1.02 s` .
The horizontal distance between the point where the first boat lands and the cliff is
`d_1= v_1 * t` , where `v_1` is the speed of the first boat, 15 m/s. Plugging in the time it takes the boat to reach the bottom, we get
`d_1 = 15 * 1.02 =15.3 m`
Similarly, we can find the distance between the point where the second boat lands and the cliff:
`d_2 = v_2*t = 26 * 1.02 =26.5 m`
So the distance between the boats when they land will be
`d_2 - d_1 = 26.5 - 15.3 = 11.2 m`
The boats will be 11.2 meters apart when they land.