Question

You throw a basketball from a cliff to the ground. The ball has an initial velocity of 12 m/s in the horizontal direction. After 3.0 seconds, the ball has a velocity of -30.0 m/s in the y-direction.

What is the direction of motion of the ball at time t= 3.0 s? (Hint: direction is related to the angle).

Answer 1

The direction of motion of the ball is approximately 68.199º below the +x semiaxis.

Step-by-step explanation:

The ball has a two-dimension motion, to be exact, a parabolical motion, that is, a motion at constant velocity in the x-direction and free fall motion in the y-motion. Hence, the horizontal velocity of the basketball is 12 meters per second and its direction (
`\theta`), in sexagesimal degrees, is calculated by this trigonometrical expression:

`\theta = \tan^(-1) (v_(y))/(v_(x))` (1)

Where:

`v_(x)` - Horizontal velocity, in meters per second.

`v_(y)` - Vertical velocity, in meters per second.

If we know that
`v_(x) = 12\,(m)/(s)` and
`v_(y) = -30\,(m)/(s)`, then the direction of motion of the ball is:

`\theta \approx 68.199^(\circ)` (below the +x semiaxis)

The direction of motion of the ball is approximately 68.199º below the +x semiaxis.