Question

You throw a ball off of a cliff at t = 0. Its motion is described by this position vector r = 40mj^ + (10m/s)ti^ - (4.9m/s^2)t^2j^ . How far from the bottom of the cliff will it hit water?

Answer 1

The ball thrown off the cliff will hit the water 20 meters from the bottom of the cliff.

Step-by-step explanation:

The position vector r = 40mj^ + (10m/s)ti^ - (4.9m/s^2)t^2j^ describes the motion of a ball thrown off a cliff. To find the distance from the bottom of the cliff where it will hit the water, we need to determine the time when the ball hits the water.

The ball hits the water when its vertical position, y, is equal to 0. In the given position vector, the y-component is -(4.9m/s^2)t^2. Setting this equal to 0 and solving for t gives us two solutions: t = 0 and t = 2s. Since the ball is thrown off the cliff initially, the positive solution, t = 2s, is the time when it hits the water.

Therefore, to find the distance from the bottom of the cliff where it hits the water, we need to find the horizontal position, x, at t = 2s. The x-component in the given position vector is (10m/s)t. Substituting t = 2s gives us x = (10m/s)(2s) = 20m.