Question

Dock jumping In the sport of dock jumping, dogs run at full speed off the end of a dock that sits a few feet above a pool of water. The winning dog is the one that lands farthest from the end of the dock. If a dog runs at 8.5 m/s (a pretty typical speed for this event) straight off the end of a dock that is 0.61 m (2 ft, a standard height) above the water, how far will the dog go before splashing down?

Answer 1

To develop this problem we will use the linear motion kinematic equations. To find the time, we will use the information given on the y axis, which will allow us to find the time it takes to complete the route. With this time and the speed given in the horizontal direction, we can calculate the total distance traveled in that direction. Consider the motion of dog along the Y-direction

$V_{0y} = \text{Initial velocity} = 0 m/s$

$a = \text{Acceleration} = 9.8m/s^2$

$y = \text{displacement} = 0.61 m$

$t = \text{time of travel}$

Using the equation,

$y = V_{0y} t + \frac{1}{2} a t^2$

$0.61 = 0*t + \frac{1}{2} (9.8) t^2$

$t = 0.35s$

Along the x-direction

$V_{0x} = \text{Constant velocity along X-direction} = 8.5 m/s$

Horizontal Displacement,

$x = V_{0x}t$

$x = (8.5)(0.35)$

$x = 2.975 m$

Therefore the dog goes before splashing down a distance of 2.975m