A football is kicked at an angle of 45° to the horizontal over a defense line up with a velocity of 15 ms^ -1…

Analyze the question: This is projectile motion. We need the horizontal velocity at the highest point of the flight, with air friction neglected.

Concept:

• With no friction, no force acts horizontally, so the horizontal component of velocity stays constant throughout the flight.
• At the highest point, only the vertical component of velocity is zero. The horizontal component is unchanged.

Solution:

Horizontal component of the initial velocity:

u_x = u cos θ = 15 × cos 45°

cos 45° = 0.7071

u_x = 15 × 0.7071 ≈ 10.6 m/s

Since horizontal velocity is constant, at the highest point:

v_x = 10.6 m/s

(At that point v_y = u sin θ − gt = 0, so the ball's speed is just v_x.)

Answer: The horizontal velocity at the highest point is about 10.6 m/s, which matches option C.

Tip: In projectile questions, remember:

• Horizontal: constant velocity, u cos θ
• Vertical: changes by gravity, u sin θ − gt
• At the top, the vertical velocity is zero, not the horizontal velocity. Option A is the common trap.