A football is kicked from the ground at 27 m/s at an angle of 42 degrees above the horizontal (ground). No one gets in the way until after it hits the ground. Use the value g=9.8 m/s^2. Assuming that the ground is level and that there is negligible air resistance, wind, etc.
a)What is the displacement of the ball (distance and direction) 1.0 s after the ball is kicked?
b)What is the velocity of the ball (speed and direction) 1.0 s after the ball is kicked?
a) Well, if we want to know the displacement of the ball 1.0 s after it is kicked, we'll need to break out some math. First, we need to find out how far horizontally the ball travels in that time. We can use the horizontal component of the initial velocity, which is given by v₀x = v₀ * cos(θ), where v₀ is the initial velocity and θ is the angle. So, v₀x = 27 m/s * cos(42°). Go ahead, bust out your calculator. Okay, got the answer? Good. Now multiply that value by the time, t = 1.0 s, to get the horizontal distance, d = v₀x * t. So, the displacement of the ball horizontally 1.0 s after it is kicked is d = (27 m/s * cos(42°)) * 1.0 s.
For the vertical displacement, it's a bit trickier. We need to find the vertical component of the initial velocity, which is given by v₀y = v₀ * sin(θ), where sin(θ) is also the ratio of the opposite side to the hypotenuse in a right triangle. So, v₀y = 27 m/s * sin(42°). Again, feel free to use your calculator. Now, we'll use a little bit of kinematics to find the vertical displacement, which is given by Δy = v₀y * t + (1/2) * g * t². Plugging in our values, we get Δy = (27 m/s * sin(42°)) * 1.0 s + (1/2) * (9.8 m/s²) * (1.0 s)².
Now to combine the horizontal and vertical displacements to get the total displacement. The total displacement, R, is simply the hypotenuse of a right triangle with the horizontal and vertical displacements as the legs. So, R = √(d² + Δy²). And there you have it!
b) The velocity of the ball 1.0 s after it is kicked is given by the magnitude of the combined horizontal and vertical velocities. The horizontal velocity, vₓ, is constant since there is no horizontal acceleration. We already calculated vₓ = v₀x = 27 m/s * cos(42°). The vertical velocity, vᵧ, changes due to the acceleration due to gravity, g. We can find vᵧ using the equation vᵧ = v₀y + g*t.
To find the magnitude of the velocity, we'll use the Pythagorean theorem. The magnitude, v, is given by v = √(vₓ² + vᵧ²). And just like that, we have the speed of the ball!
Now, go forth and calculate! Don't forget to get back to me with the results. I'll be here, amusing myself with silly jokes while you crunch numbers. Good luck!
a) To find the displacement of the ball 1.0 s after it is kicked, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component (Vx) remains constant throughout the motion and is given by:
Vx = V * cos(theta)
where V is the initial velocity (27 m/s) and theta is the angle of projection (42 degrees).
Vx = 27 m/s * cos(42 degrees)
Vx ≈ 20.360 m/s
The vertical component (Vy) of velocity changes due to the acceleration due to gravity. At t = 1.0 s, we can use the following equation to find the vertical displacement:
Vy = Vy0 + gt
where Vy0 is the initial vertical velocity (Vy0 = V * sin(theta)), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time (1.0 s).
Vy = V * sin(theta) + g * t
Vy ≈ 27 m/s * sin(42 degrees) + (-9.8 m/s^2) * 1.0 s
Vy ≈ 17.704 m/s - 9.8 m/s
Vy ≈ 7.904 m/s
Now, we can find the vertical displacement (sy) using the equation:
sy = sy0 + Vyt + (1/2)gt^2
where sy0 is the initial vertical displacement (0 m), Vyt is the initial vertical velocity multiplied by time (7.904 m/s * 1.0 s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time (1.0 s).
sy = 0 + 7.904 m/s * 1.0 s + (1/2) * (-9.8 m/s^2) * (1.0 s)^2
sy ≈ 7.904 m - 4.9 m
sy ≈ 2.004 m
The displacement of the ball 1.0 s after it is kicked is approximately 20.360 m horizontally and 2.004 m vertically. To find the total distance, we can use the Pythagorean theorem:
distance = √(sx^2 + sy^2)
distance ≈ √((20.360 m)^2 + (2.004 m)^2)
distance ≈ √(414.6256 m^2 + 4.016032 m^2)
distance ≈ √(418.641632 m^2)
distance ≈ 20.46 m
Therefore, the displacement of the ball 1.0 s after it is kicked is approximately 20.46 m at an angle of 5.52 degrees above the horizontal (ground).
b) The velocity of the ball at any given time can be found by combining the horizontal and vertical components of velocity.
The horizontal velocity remains constant throughout the motion and is given by Vx = V * cos(theta), as mentioned before:
Vx ≈ 20.360 m/s
The vertical velocity (Vy) changes due to the acceleration due to gravity. At t = 1.0 s, we calculated Vy to be approximately 7.904 m/s.
To find the total velocity (V) and its direction (theta), we can again use the Pythagorean theorem and trigonometry:
V = √(Vx^2 + Vy^2)
V = √((20.360 m/s)^2 + (7.904 m/s)^2)
V ≈ √(414.6256 m^2/s^2 + 62.419216 m^2/s^2)
V ≈ √(476.044816 m^2/s^2)
V ≈ 21.81 m/s
The direction (theta) can be found using the inverse tangent function:
theta = atan(Vy/Vx)
theta = atan(7.904 m/s / 20.360 m/s)
theta ≈ 21.35 degrees
Therefore, the velocity of the ball 1.0 s after it is kicked is approximately 21.81 m/s at an angle of 21.35 degrees above the horizontal (ground), directed towards the upper-right quadrant.
To find the displacement and velocity of the ball 1.0 s after it is kicked, we can use the equations of projectile motion. Here's how you can solve it step by step:
Step 1: Break down the initial velocity into its horizontal and vertical components.
The horizontal component (Vx) can be found using Vx = V0 * cos(theta), where V0 is the initial velocity and theta is the angle above the horizontal.
The vertical component (Vy) can be found using Vy = V0 * sin(theta).
In this case, V0 = 27 m/s and theta = 42 degrees. Plugging these values into the equations, we find:
Vx = 27 m/s * cos(42 degrees)
Vy = 27 m/s * sin(42 degrees)
Step 2: Calculate the time of flight (T) for the projectile. This is the time it takes for the ball to hit the ground. The formula is: T = (2 * Vy) / g, where g is the acceleration due to gravity.
Plugging in the values, we get:
T = (2 * Vy) / g
Step 3: Calculate the distance (Sx) traveled horizontally after 1.0 second. The formula is: Sx = Vx * t, where t is the time.
Plugging in the values, we get:
Sx = Vx * 1.0 s
Step 4: Calculate the vertical displacement (Sy) after 1.0 second. The formula is: Sy = Vy * t - (0.5 * g * t^2)
Plugging in the values, we get:
Sy = Vy * 1.0 s - (0.5 * g * (1.0 s)^2)
Step 5: Calculate the magnitude and direction of the displacement. The magnitude (S) can be found using the Pythagorean theorem: S = sqrt(Sx^2 + Sy^2). The direction can be found using the inverse tangent function: theta = arctan(Sy / Sx).
Step 6: Calculate the magnitude and direction of the velocity. The magnitude (V) can be found using V = sqrt(Vx^2 + Vy^2). The direction can be found using the inverse tangent function: alpha = arctan(Vy / Vx).
Now let's perform the calculations:
Step 1:
Vx = 27 m/s * cos(42 degrees)
Vy = 27 m/s * sin(42 degrees)
Step 2:
T = (2 * Vy) / g
Step 3:
Sx = Vx * 1.0 s
Step 4:
Sy = Vy * 1.0 s - (0.5 * g * (1.0 s)^2)
Step 5:
S = sqrt(Sx^2 + Sy^2)
theta = arctan(Sy / Sx)
Step 6:
V = sqrt(Vx^2 + Vy^2)
alpha = arctan(Vy / Vx)
Performing these calculations will give you the answers to part (a) and (b) of the question.