A soccer ball is kicked with an initial speed of 10.8 m/s in a direction 24.9° above the horizontal. Calculate the magnitude of its velocity 0.310 s after being kicked. What was its direction? Calculate the magnitude of its velocity 0.620 s after being kicked. What was its angle relative to the horizontal (choose positive for above and negative for below)?
See 11:25 Am post.
To calculate the magnitude of velocity at a specific time after being kicked, we need to split the initial velocity into its horizontal and vertical components and then calculate the magnitude at the desired time.
Let's start by finding the initial horizontal and vertical components of the velocity.
Horizontal component (Vx):
Vx = V * cos(θ)
where V is the initial speed and θ is the angle above the horizontal.
Vx = 10.8 m/s * cos(24.9°)
Vx ≈ 9.748 m/s
Vertical component (Vy):
Vy = V * sin(θ)
where V is the initial speed and θ is the angle above the horizontal.
Vy = 10.8 m/s * sin(24.9°)
Vy ≈ 4.699 m/s
Now let's calculate the magnitude of velocity at 0.310 s after being kicked.
We know that the horizontal velocity remains constant because there are no horizontal forces acting on the ball. The vertical velocity, however, changes due to the acceleration of gravity.
Horizontal velocity (Vx) remains the same:
Vx = 9.748 m/s
Vertical velocity (Vy) changes due to gravity:
Vy = Vy(initial) + (acceleration due to gravity) * time
where Vy(initial) is the initial vertical velocity, acceleration due to gravity is approximately 9.8 m/s^2, and time is the desired time (0.310 s).
Vy = 4.699 m/s + (9.8 m/s^2) * 0.310 s
Vy ≈ 7.509 m/s
To find the magnitude of velocity at 0.310 s after being kicked, we can use the Pythagorean theorem:
Magnitude of velocity (V) = √(Vx^2 + Vy^2)
V ≈ √((9.748 m/s)^2 + (7.509 m/s)^2)
V ≈ √(95.050 m^2/s^2)
V ≈ 9.751 m/s
So, the magnitude of the velocity 0.310 s after being kicked is approximately 9.751 m/s.
To calculate the direction of the velocity, we need to find the angle it makes with the horizontal. We can use the inverse tangent function to find this angle.
Angle (θ) = arctan(Vy / Vx)
θ = arctan(7.509 m/s / 9.748 m/s)
θ ≈ 39.994°
Since the angle is above the horizontal, the direction of the velocity 0.310 s after being kicked is approximately 39.994° (rounded to the nearest degree) above the horizontal.
To calculate the magnitude of velocity at 0.620 s after being kicked, we can repeat the process.
Horizontal velocity (Vx) remains the same:
Vx = 9.748 m/s
Vertical velocity (Vy) changes due to gravity:
Vy = Vy(initial) + (acceleration due to gravity) * time
where Vy(initial) is the initial vertical velocity, acceleration due to gravity is approximately 9.8 m/s^2, and time is the desired time (0.620 s).
Vy = 4.699 m/s + (9.8 m/s^2) * 0.620 s
Vy ≈ 10.306 m/s
Magnitude of velocity (V) = √(Vx^2 + Vy^2)
V ≈ √((9.748 m/s)^2 + (10.306 m/s)^2)
V ≈ √(190.090 m^2/s^2)
V ≈ 13.793 m/s
So, the magnitude of the velocity 0.620 s after being kicked is approximately 13.793 m/s.
To find the angle it makes with the horizontal, we can use the inverse tangent function.
Angle (θ) = arctan(Vy / Vx)
θ = arctan(10.306 m/s / 9.748 m/s)
θ ≈ 49.857°
Since the angle is above the horizontal, the angle relative to the horizontal is approximately 49.857° (rounded to the nearest degree) above.