Astone is thrown at an angle of 45 degrees above the horizontal from the top edge of a cliff with an initial speed of 14 m/s. a stop watch measures the stone's trajectory time from top of cliff to bottom to be 6.9 s. how far (in meter) out from the cliff's edge does the stone travel horizontally? (air resistance is negligible)

Stone fall far from the cliff's edge horizontally is 68.31 m

Explanation:

Given data

Angle = ∅ = 45°

Initial velocity = v₀ = 14 ms⁻¹

Time = t = 6.9 s

As the stone thrown at an angle of 45° above horizontal, so it follow projectile path.

As we know that in projectile motion horizontal component of velocity throughout remain same. So there is no acceleration along horizontal. So we can use simple formula

                  Distance = Velocity × Time

                             Δx = v₀ Cos∅ × t

                             Δx = 14 Cos(45) × 6.9    

                            Δx = 68.31 m

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