A football is kicked at an angle of 45° to the horizontal over a defence line up with a velocity of 15m/s. Calculate the magnitude of the horizontal velocity of the ball at its highest point neglecting friction
the horizontal velocity is constant regardless of height
... the vertical velocity changes due to gravity
horizontal velocity = 15 m/s * cos(45º)
Yes
To calculate the magnitude of the horizontal velocity of the ball at its highest point, we can use the formula for horizontal velocity:
Vx = V * cos(θ)
where:
Vx is the horizontal velocity
V is the initial velocity (15 m/s)
θ is the launch angle (45°)
Using this formula, we can substitute the given values and calculate the magnitude of the horizontal velocity:
Vx = 15 m/s * cos(45°)
To calculate cos(45°), we can use the fact that cos(45°) = √2 / 2:
Vx = 15 m/s * (√2 / 2)
Simplifying this expression:
Vx = 15 m/s * 0.707
Vx ≈ 10.606 m/s
Therefore, the magnitude of the horizontal velocity of the ball at its highest point is approximately 10.606 m/s.
To calculate the magnitude of the horizontal velocity of the ball at its highest point, we can use trigonometric functions and the given information. Here's how you can figure it out:
1. Break down the initial velocity into its horizontal and vertical components. Since the ball is kicked at an angle of 45° to the horizontal, the initial velocity can be split equally between the horizontal and vertical directions.
Horizontal velocity (Vx) = initial velocity * cos(angle)
= 15 m/s * cos(45°)
= 10.61 m/s
Vertical velocity (Vy) = initial velocity * sin(angle)
= 15 m/s * sin(45°)
= 10.61 m/s
2. At the highest point of the ball's trajectory, the vertical velocity will be zero because the ball will momentarily stop before it starts falling back down. Therefore, the horizontal velocity at the highest point will remain constant.
Thus, the magnitude of the horizontal velocity at the ball's highest point is 10.61 m/s.