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

Answer:

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

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:

$\theta = \tan^{-1} \frac{v_{y}}{v_{x}}$ (1)

Where:

$v_{x}$ - Horizontal velocity, in meters per second.

$v_{y}$ - Vertical velocity, in meters per second.

If we know that $v_{x} = 12\,\frac{m}{s}$ and $v_{y} = -30\,\frac{m}{s}$, 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.