Question

A person kicks a ball with an initial velocity of 5.2 m/s at an angle of 28 degrees with the horizontal. The ball has an initial vertical velocity of 2.3 m/s and a total time of flight of 2 seconds. What is the maximum height reached by the ball?

A) 1.5 m.
B) 2.0 m.
C) 2.5 m.
D) 3.0 m.

Answer 1

The maximum height reached by the ball using the provided vertical velocity and the formula h = vy02 / (2 * g) is approximately 0.2699 meters. However, this does not match the given answer choices, indicating a possible error in the values provided in the question.

Step-by-step explanation:

To find the maximum height reached by the ball, we can use the formula for the vertical motion of a projectile. The vertical motion is independent of the horizontal motion and is affected by gravity. Considering the initial vertical velocity (vy0 = 2.3 m/s) and the acceleration due to gravity (g = -9.81 m/s2), the maximum height (h) can be calculated using the following equation:

h = vy02 / (2 * g)

Substituting the given values:

h = (2.3)^2 / (2 * -9.81)

This gives:

h = 5.29 / 19.62

h = 0.2699 meters

However, since none of the answer choices match, there might be an error in the problem statement or the values provided. If we are to believe the time of flight and vertical velocity given, the method above would yield the correct approach to the maximum height calculation in a vacuum without air resistance.