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)?
Vo = 10.8m/s[24.9o]
Xo = 10,8*cos24.9 = 9.8 m/s
Yo = 10.8*sin24.9 = 4.55 m/s
a. Y = Yo + g*t
Y = 4.55 + (-9.8)*0.310 = 1.51 m/s.
V^2 = Xo^2 + Y^2
V^2 = 9.8^2 + 1.51^2 = 98.3
V = 9.9 m/s.
b. = tan A = Y/Xo = 1.51/9.8 = 0.15408
A = 8.76o = The direction.
c. Y = Yo + g*t
Y = 4.55 + (-9.8)*0.620 = -1.53 m/s.
V^2 = Xo^2 + Y^2
V^2 = 9.8^2 + (-1.53^2) = 98.38
V = 9.9 m/s.
tan B = Y/Xo = -1.53/9.8 = -0.15612
B = -8.87o = The direction.
NOTE: The negative velocity(-1.53) means that the ball is falling and each
point during the rise time is repeated during the fall time.
To calculate the magnitude of the velocity of the soccer ball at a specific time, we need to break down the initial velocity into its horizontal and vertical components.
1. Finding the horizontal component:
The horizontal component of the soccer ball's velocity remains constant since there are no horizontal forces acting on it. The horizontal component can be found using the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of velocity, V is the initial velocity, and θ is the angle of the velocity vector above the horizontal.
V = 10.8 m/s (given)
θ = 24.9° (given)
Vx = 10.8 * cos(24.9°)
Vx ≈ 9.725 m/s (rounded to three decimal places)
2. Finding the vertical component:
The vertical component of the soccer ball's velocity changes due to the acceleration due to gravity. The vertical component can be found using the formula:
Vy = V * sin(θ) - g * t
where Vy is the vertical component of velocity, V is the initial velocity, θ is the angle of the velocity vector above the horizontal, g is the acceleration due to gravity (approximately -9.8 m/s²), and t is the time.
V = 10.8 m/s (given)
θ = 24.9° (given)
g = -9.8 m/s² (acceleration due to gravity)
t = 0.31 s (first time)
Vy = 10.8 * sin(24.9°) - 9.8 * 0.31
Vy ≈ 6.979 m/s (rounded to three decimal places)
3. Finding the magnitude of the velocity at 0.31 s:
To find the magnitude of the velocity, we can use the Pythagorean theorem since the horizontal and vertical components are perpendicular.
Magnitude of velocity (V) = √(Vx^2 + Vy^2)
Magnitude of velocity (V) = √(9.725^2 + 6.979^2)
Magnitude of velocity (V) ≈ √(95.003 + 48.646)
Magnitude of velocity (V) ≈ √143.649
Magnitude of velocity (V) ≈ 11.99 m/s (rounded to two decimal places)
4. Finding the direction at 0.31 s:
The direction can be found using the inverse tangent function (tan^(-1)) of the ratio of the vertical component to the horizontal component.
Direction = tan^(-1)(Vy / Vx)
Direction = tan^(-1)(6.979 / 9.725)
Direction ≈ 37.220° (rounded to three decimal places)
Repeat steps 2-4 to calculate the magnitude and direction at 0.620 seconds.