Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball\'s trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligable, and plot the horizontal and vertical components of the ball\'s velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take \"north\" and \"up\" as the positive x and y directions, respectively, and use g ≈10 m/s2 for the downward acceleration due to gravity.
Horizontal Velocity: a horizontal line at 10 m/s
Vertical Velocity: a straight line, starting at 10 m/s at 0 sec, decreases 5 m/s every 0.5 second (or 10 m/s each 1 second).
Well, Emmy certainly knows how to kick a ball! Let's break this down.
First, let's calculate the initial velocity of the ball. Since the ball is kicked at an angle of 45°, we can use trigonometry (specifically, the sine and cosine functions) to find the horizontal and vertical components of the initial velocity.
The horizontal component of the initial velocity (Vx) will remain constant throughout the motion because there is no horizontal acceleration. So, we can simply use the formula Vx = V0 * cos(θ), where V0 is the initial velocity and θ is the launch angle.
For the vertical component of the initial velocity (Vy), we need to consider that the ball is being kicked upwards. So, Vy will be positive and we can use the formula Vy = V0 * sin(θ).
Since we are given the angle as 45°, we can simplify the equations to Vy = Vx = V0 / √2.
Next, let's consider the vertical motion of the ball. Here, we can use the formula h = Vy0 * t - (1/2) * g * t^2, where h is the vertical distance, Vy0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Given that the ball lands after 2 seconds, we can substitute t = 2 into the equation and solve for h to find the vertical distance traveled.
So, h = Vy0 * 2 - (1/2) * 10 * 2^2 = Vy0 * 2 - 20
Since we are given that the ball lands 20 meters away to the north, we can infer that the horizontal distance (Dx) traveled by the ball is 20 meters.
Now, let's calculate the horizontal component of the velocity at any given time. Since there is no horizontal acceleration, the horizontal component of the velocity remains constant throughout the motion. So, Dx = Vx * t.
Since we know that Vx = V0 / √2, we can substitute it into the equation to find Dx = (V0 / √2) * t.
Finally, let's put it all together to plot the horizontal and vertical components of the ball's velocity as a function of time.
Horizontal component (Vx): Constant throughout the motion, represented by a horizontal line.
Vertical component (Vy): Initially positive and decreases due to the acceleration due to gravity, represented by a decreasing curve.
I hope this explanation doesn't fall flat! Let me know if you have any more questions.
To determine the horizontal and vertical components of the ball's velocity as a function of time, we can break down the motion into its x and y components.
Given information:
- Angle of 45°
- Time of flight: 2 seconds
- Displacement in the x-direction (horizontal): 20 m
- Acceleration due to gravity (g): 10 m/s^2
Step 1: Calculate the initial velocity components.
Since the angle of launch is 45°, the initial velocity can be divided equally between the x and y components.
Let the initial velocity of the ball be denoted as 'V0'.
V0 = V0x = V0y
Step 2: Calculate the initial velocity components.
We can calculate the initial velocity by using the time of flight and the displacement in the x-direction.
Displacement in the x-direction (horizontal) = V0x * time
20 m = V0 * cos(45°) * 2
V0 * cos(45°) = 20 m / 2
V0 * cos(45°) = 10 m/s
V0 = 10 m/s / cos(45°)
V0 = 14.14 m/s
Step 3: Calculate the y-component of velocity.
The vertical component of velocity is affected by gravity. The initial velocity in the y-direction can be calculated using the formula:
V0y = V0 * sin(45°)
V0y = 14.14 m/s * sin(45°)
V0y = 10 m/s
Step 4: Plot the horizontal and vertical components of the ball's velocity as a function of time.
Using the calculated initial velocity components, we can now plot the horizontal and vertical components of the ball's velocity as a function of time during the ball's flight.
Horizontal component (Vx) remains constant throughout the motion: Vx = V0x = 14.14 m/s
Vertical component (Vy) changes due to the effect of gravity: Vy = V0y - g * t
where,
- Vy is the vertical component of the velocity at time 't',
- V0y is the initial vertical component of velocity,
- g is acceleration due to gravity (10 m/s^2),
- t is the time.
Thus, the components of the ball's velocity as a function of time can be given as follows:
Horizontal component: Vx = 14.14 m/s (constant)
Vertical component: Vy = 10 m/s - 10 m/s^2 * t
Now, you can plot the horizontal and vertical components of the ball's velocity as a function of time.
To solve this problem, we need to break down the initial velocity of the soccer ball into its horizontal and vertical components.
We know that the angle of the ball's trajectory is 45°, and we can assume that the initial velocity is the same for both components.
The horizontal component of the velocity (Vx) remains constant throughout the ball's flight because there is no acceleration in that direction. The vertical component of the velocity (Vy) changes due to the acceleration due to gravity.
To find the initial velocity (Vo), we can use the formula:
Vo = V / sin(θ)
where V is the initial velocity of the soccer ball and θ is the angle of the trajectory. In this case, V is the total initial velocity, which we are trying to find.
Using the given information, the time of flight (t) is 2 seconds, and the horizontal displacement (Δx) is 20 meters.
First, let's find the initial velocity (Vo):
Vo = V / sin(θ)
Vo = (Δx) / (t * sin(θ))
Vo = 20 / (2 * sin(45°))
Vo ≈ 20 / (2 * 0.7071)
Vo ≈ 20 / 1.4142
Vo ≈ 14.14 m/s
Now that we know the initial velocity, we can find the horizontal and vertical components of the velocity as a function of time.
The horizontal component of the velocity (Vx) is constant and given by:
Vx = Vo * cos(θ)
Vx = 14.14 * cos(45°)
Vx ≈ 14.14 * 0.7071
Vx ≈ 10 m/s
The vertical component of the velocity (Vy) changes due to the acceleration due to gravity. We can use the equation of motion:
Vy = Vo * sin(θ) - gt
where g is the acceleration due to gravity (approximately 10 m/s²) and t is the time.
Plugging in the values, we get:
Vy = 14.14 * sin(45°) - 10 * t
Vy ≈ 14.14 * 0.7071 - 10 * t
Vy ≈ 10 - 10t
So, the horizontal component of the velocity (Vx) remains constant at 10 m/s, while the vertical component of the velocity (Vy) changes linearly with time, starting at 10 m/s and decreasing at a rate of 10 m/s².