Question

Football is kicked at an angle of 45° to the horizontal over a defense line up with a velocity of 15ms⁻¹ Calculate the magnitudeof horizontal velocity of the ball at its highest point [Neglect friction, g = 10ms⁻²].

Answer 1

The magnitude of the horizontal velocity of a football kicked at a 45° angle to the horizontal with an initial velocity of 15 m/s, neglecting air resistance, is approximately 10.61 m/s.

Step-by-step explanation:

To calculate the magnitude of the horizontal velocity of a football kicked at a 45° angle to the horizontal, we can use the initial velocity and trigonometry. The horizontal velocity component (Vx) can be calculated using the cos function, as follows:

Vx = V * cos(θ)

where V is the initial velocity of the ball and θ is the launch angle. Given that the velocity (V) is 15 m/s and the angle (θ) is 45°, we find:

Vx = 15 m/s * cos(45°)

Since cos(45°) = 0.7071 (approximately), the horizontal velocity at the highest point (which remains constant due to the neglect of air resistance) is:

Vx ≈ 15 m/s * 0.7071 ≈ 10.61 m/s.