A ball is dropped from rest from the top of a tall cliff. In the last 1 s of its fall before hitting the ground, it covers a distance of 45 m . What is the total height of the cliff? (Take g = 10 m s ⁻² )

Step-by-step solution

Let the velocity of the ball 1 s before hitting the ground be u . For the last 1 s of the fall, the distance covered is 45 m . Using the second equation of motion: s = ut + 1 2 gt^2 45 = u(1) + 1 2 (10)(1)^2 45 = u + 5 u = 40 m s ⁻¹ This velocity u = 40 m s ⁻¹ is attained by the ball after falling from rest. Let the time taken to reach this velocity be t₁ . Using v = u_ initial + gt₁ : 40 = 0 + 10t₁ t₁ = 4 s The total time of fall is the time taken to reach 40 m s ⁻¹ plus the last 1 s : t_ total = 4 + 1 = 5 s The t