Question

run at full speed off the end of pool of water. The winning dog is the one that lands farthest from the end of the dock. If a 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? In the sport of dock jumping, dogs a dock that sits a few feet above a dog

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the dog = V₀ = 8.5 m/s

The dog runs straight off the end of the dock that is the velocity of the dog is horizontal when it leaves the dock.

Initial horizontal velocity of the dog = V_x0 = 8.5 m/s

Initial vertical velocity of the dog = V_y0 = 0 m/s

Height of the dock = H= 0.61 m

Time before the dog hits the water = T

The displacement is downwards for the dog hence it is negative.

-H = V_y0T + gT²/2

-0.61 = (0)T + (-9.81)T²/2

T = 0.352 sec

There is no horizontal force acting on the dog hence the velocity of the dog in the horizontal direction remains constant.

Horizontal distance traveled by the dog = R

R = V_x0T

R = (8.5)(0.352)

R = 2.99 m

The dog will go 2.99 m before splashing in the water.