Question

A player hits a tennis ball in the air with an initial velocity of 32 meters per second at 35 degrees from the vertical how fast is the ball moving at the highest point

Answer 1

Answer: 18.35 m/s

Explanation:

At the highest point of trajectory, the vertical component of the velocity would be zero and the tennis ball would have horizontal component of velocity.

It is given that the initial velocity of the ball is 32 m/s and it makes 35° with the vertical. Hence the horizontal component of the velocity,

v sin θ = 32 m/s × sin 35° = 18.35 m/s

Hence, at the highest point in its trajectory, the tennis ball would be moving with the speed  18.35 m/s.

Answer 2

Answer:

26.2 m/s

Explanation:

Let's start by calculating the components of the initial velocity in the horizontal and vertical directions. The angle with the vertical is 35 degrees, so the angle with the horizontal is 90-35=55 degrees. Therefore:

- Horizontal: $v_x = v cos \theta = (32 m/s)(cos 55^{\circ})=18.4 m/s$

- Vertical: $v_y = v sin \theta =(32 m/s)(sin 55^{\circ})=26.2 m/s$

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal uniform motion, with constant velocity (18.4 m/s), because the acceleration in this direction is zero (there are no forces acting on the ball along the horizontal direction)

- A vertical accelerated motion, with constant acceleration $g=-9.8 m/s^2$ in the downward direction due to the force of gravity

As the ball moves upward, its vertical velocity decreases (due to the acceleration of gravity), until it becomes zero at the highest point of the ball. The horizontal velocity, however, remains constant. Therefore, at the highest point the ball will only have horizontal velocity: therefore, its velocity will be

18.4 m/s