Speed of a Tennis Ball Launching at an Angle of 45 Degrees

What is the speed of the tennis ball as it leaves the launcher?

Can we determine the speed of the tennis ball based on its initial velocity, launch angle, and other factors provided?

Speed of the Tennis Ball as it Leaves the Launcher

The speed of the tennis ball as it leaves the launcher is approximately 12.7 meters per second.

To determine the speed of the tennis ball as it leaves the launcher, we can analyze its motion in the vertical and horizontal directions separately. Since there is no air resistance and the ball lands at the same height as it was launched, we can assume the vertical velocity component remains constant throughout the motion.

Given that the ball reaches its maximum height 0.92 seconds after launch, we can use the kinematic equation for vertical motion: vf = vi + gt, where vf is the final vertical velocity, vi is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s²), and t is the time taken. Substituting the values, we have 0 = 9.0 m/s - 9.8 m/s² × 0.92 s, which allows us to solve for the initial vertical velocity.

Next, we can find the initial horizontal velocity component. Since the launch angle is 45 degrees and the initial horizontal velocity is equal to the initial vertical velocity, we can use trigonometry to determine its value.

Finally, we can calculate the overall speed of the ball as it leaves the launcher by combining the vertical and horizontal velocity components using the Pythagorean theorem.

By performing the calculations, the speed of the tennis ball as it leaves the launcher is approximately 12.7 meters per second.