A player kicks a football with an initial speed of 25 ms⁻¹ at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g = 10 ms⁻²)

Solution

Physics sir · 3 steps

At the top only the vertical velocity becomes zero, so work with the vertical component alone.

1. Vertical component. Only the vertical part of the velocity is slowed by gravity: u_y = u sin 45° = 25/√2 ≈ 17.68 m s⁻¹. At the highest point u_y has dropped to zero.
2. Time to the top. From 0 = u_y - gt, the time is t = (u sin θ)/g = 17.68/10 ≈ 1.77 s.

   hₘₐₓ = (u² sin² θ)/2g = (625 × 1/2)/(2 × 10) = 15.625 m

3. So. The football rises to 15.625 m and takes 1.77 s to get there, which is option (c).

The idea to keep. For the highest point, use only u sin θ: time up is (u sin θ)/g and hₘₐₓ = (u²sin² θ)/2g. The full flight takes twice the time up.

Answer from the NTA final answer key. Solution by PrepWiser.