Question

A stone is dropped from a high altitude and 3 s later another stone is projected downward with a speed of 150 ft/s. When and where will the second stone overtake the first? Ans. (a) 5.70 s (b) 520 ft

Answer 1

The first is falling freely (from a zero speed), its distance after time t is:

d₁ =  gt²/2,  where g = 9.8 m/s²

The second is falling after t₀ = 3 s  from v₀ = 150 ft/s = 45.72 m/s, its distance after time t - t₀ is:

d₂ = v₀(t - t₀) + g(t - t₀)²/2

The stones will overcome the same distance until meeting:

d₁ = d₂gt²/2 = v₀(t - t₀) + g(t - t₀)²/2

gt² = 2v₀(t - t₀) + g(t - t₀)²

gt² = 2v₀t - 2v₀t₀ + gt² - 2gt₀t + gt₀²

2(v₀ - gt₀)t = 2v₀t₀ - gt₀²

d₁ = 9.8 × 5.7² / 2 = 159.2 m = 522.3 ft

Answer:

The stones are meeting after 5.7 s of beginning flying of first stone or after 5.7 - 3 = 2.7 s of a second. Both stones overcome 522.3 ft distance.