Final answer:
The ball will reach a maximum height of approximately 7.35 meters. It will be in the air for approximately 1.22 seconds. The ball will travel a distance of approximately 39.04 meters.
Step-by-step explanation:
To find out how high the ball will go, we can analyze the vertical component of its motion. The initial velocity in the vertical direction is given by 12 m/s. Using this information, we can apply the kinematic equation:
Δy = V0yt + 0.5gt²
Where Δy is the displacement in the vertical direction (unknown), V0y is the initial velocity in the vertical direction (12 m/s), t is the time (unknown), and g is the acceleration due to gravity (-9.8 m/s²). By rearranging the equation, we can solve for t:
t = (V0y ± √(V0y² - 2gΔy)) / g
Since we are interested in the time it takes for the ball to reach its maximum height, we can set the displacement Δy equal to zero:
t = V0y / g
Now we can substitute the given values to find the time:
t = 12 m/s / 9.8 m/s² = 1.22 s
So, it takes approximately 1.22 seconds for the ball to reach its maximum height.
To determine the maximum height, we can use the equation:
Vf² = V0² + 2aΔy
Where Vf is the final velocity (which is zero at the maximum height), V0 is the initial velocity in the vertical direction (12 m/s), a is the acceleration due to gravity (-9.8 m/s²), and Δy is the displacement (unknown). By rearranging the equation, we get:
Δy = (Vf² - V0²) / (2a)
Plugging in the values, we have:
Δy = (0 - 12 m/s)² / (2 * -9.8 m/s²) = 7.35 m
Therefore, the maximum height that the ball will reach is approximately 7.35 meters.
To find out how long the ball will be in the air, we can analyze the horizontal component of its motion. The initial velocity in the horizontal direction is given by 16 m/s. The time of flight can be determined using the equation:
t = Δx / V0x
Where t is the time of flight (unknown), Δx is the range (unknown), and V0x is the initial velocity in the horizontal direction (16 m/s). Since we are interested in the range, we can rewrite the equation as:
Δx = V0x * t
Plugging in the values, we have:
Δx = 16 m/s * (2 * 1.22 s) = 39.04 m
Therefore, the ball will travel approximately 39.04 meters.