A ball is thrown horizontally from a cliff with an initial velocity of 15 m/s. If the cliff is 45 m high, how far from the base of the cliff will the ball land?

Core Concept

First, find the time it takes for the ball to fall 45 m using y = y₀ + v₀t + (1/2)at². For vertical motion: 0 = 45 + 0t - (1/2)(9.8)t², which gives t = √(90/9.8) ≈ 3.02 s. The horizontal distance is x = v₀t = 15 × 3.02 ≈ 45.3 m, which rounds to 45 m.