To tackle this problem, we need to break it down into three parts: finding the maximum height the ball will reach, calculating the total horizontal distance it travels, and determining its velocity just before it hits the ground. Let's go through each part step by step.
Maximum Height Calculation
The ball is thrown with an initial velocity that has both horizontal and vertical components. The vertical component of the velocity is crucial for determining how high the ball will rise. In this case, the vertical component is given as 6.1 m/s.
To find the maximum height, we can use the following kinematic equation:
v² = u² + 2as
- v = final velocity (0 m/s at the maximum height)
- u = initial vertical velocity (6.1 m/s)
- a = acceleration due to gravity (-9.81 m/s², negative because it acts downward)
- s = displacement (the additional height gained)
Rearranging the equation to solve for s gives us:
s = (v² - u²) / (2a)
Substituting the values:
s = (0 - (6.1)²) / (2 * -9.81)
s = -37.21 / -19.62 = 1.90 m
Now, to find the total maximum height, we add this additional height to the height at which the velocity was observed:
Total Height = 9.11 m + 1.90 m = 11.01 m
Horizontal Distance Traveled
The horizontal distance traveled can be calculated using the time the ball is in the air and the horizontal component of the velocity. The horizontal component is given as 7.6 m/s.
First, we need to find the total time the ball is in the air. The time to reach the maximum height can be calculated using:
t = u / a
Where:
- u = initial vertical velocity (6.1 m/s)
- a = acceleration due to gravity (9.81 m/s²)
Substituting the values:
t = 6.1 / 9.81 ≈ 0.62 s
This is the time to reach the maximum height. The total time of flight is double this, as the time taken to ascend is equal to the time taken to descend:
Total Time = 2 * 0.62 s ≈ 1.24 s
Now, we can calculate the horizontal distance:
Horizontal Distance = Horizontal Velocity * Total Time
Horizontal Distance = 7.6 m/s * 1.24 s ≈ 9.42 m
Velocity Before Hitting the Ground
To find the velocity just before the ball hits the ground, we need to consider both the horizontal and vertical components of the velocity at that moment. The horizontal component remains constant at 7.6 m/s. The vertical component can be calculated using the following kinematic equation:
v = u + at
Where:
- u = initial vertical velocity (6.1 m/s)
- a = acceleration due to gravity (9.81 m/s²)
- t = total time of flight (1.24 s)
Substituting the values:
v = 6.1 + (9.81 * 1.24) ≈ 6.1 + 12.17 ≈ 18.27 m/s
Now, we can find the magnitude of the total velocity just before hitting the ground using the Pythagorean theorem:
Velocity Magnitude = √(horizontal² + vertical²)
Velocity Magnitude = √((7.6)² + (18.27)²)
Velocity Magnitude = √(57.76 + 333.84) ≈ √391.6 ≈ 19.8 m/s
In summary, the answers to your questions are:
- Maximum Height: 11.01 m
- Total Horizontal Distance: 9.42 m
- Velocity Before Hitting the Ground: 19.8 m/s
