Question

What is the distance from the bottom of the cliff where the ball will hit the water, given the motion described by the position vector r=10mj+(5m/s)ti − (4.9m/s²)t²j?

Answer 1

The distance from the bottom of the cliff where the ball will hit the water is 5sqrt(2) meters.

Step-by-step explanation:

The given position vector describes the motion of a ball thrown horizontally off a cliff.

To find the distance from the bottom of the cliff where the ball will hit the water, we need to determine the time it takes for the ball to hit the water.

Using the given equation for the y-coordinate of the position vector, -4.9t^2, and setting it equal to zero, we can solve for the time. Solving for t gives us:

t = 0 or t = sqrt(2).

Since t = 0 represents the starting point of the motion, we ignore it. Therefore, the ball will hit the water after sqrt(2) seconds.

To find the distance from the bottom of the cliff, we can substitute the value of t into the x-coordinate of the position vector. The x-coordinate is given by 5t. So, the distance is 5(sqrt(2)) = 5sqrt(2) meters.