Final answer:
Using the kinematic equation and given the initial velocity of 15 m/s and gravity 9.8 m/s², the ball will reach a height of approximately 11.48 meters before it starts to fall.
Step-by-step explanation:
To calculate how high the ball will go before it begins to fall, we can use the kinematic equation for an object under constant acceleration due to gravity (g = 9.8 m/s²). We'll use the following equation:
vf² = vi² + 2gd
where vf is the final velocity (0 m/s at the highest point), vi is the initial velocity (15 m/s), g is the acceleration due to gravity, and d is the distance the ball will travel upwards (the height we are looking to find).
Using this equation, we get:
0 = (15 m/s)² + 2(-9.8 m/s²)(d)
Plugging in the values, the equation becomes:
0 = (15 m/s)^2 + 2(-9.8 m/s^2)d
Solving for d, we get:
d = (15 m/s)^2 / (2 * 9.8 m/s^2)
d = 11.47 m
Therefore, the ball will go approximately 11.47 meters high before it begins to fall.
Solving for d, we find:
d = (15 m/s)² / (2 * 9.8 m/s²) = 11.48 m
Therefore, the ball will reach a height of approximately 11.48 meters before beginning to fall.