A ball is projected vertically upward with an initial velocity of 50 m s⁻¹ at t = 0 s. At t = 2 s, another ball is projected vertically upward with same velocity. At t = s, second ball will meet the first ball (g = 10 m s⁻²).

Question

Accepted Answer

6. y₁ = 50t - 5t²; y₂ = 50(t - 2) - 5(t - 2)² for t ≥ 2. Meet when y₁ = y₂: 50t - 5t² = 50t - 100 - 5t² + 20t - 20 ⇒ 0 = 20t - 120. t = 6 s (each ball is then 3 s from its own launch, one going up, one coming down, by symmetry).

A ball is projected vertically upward with an initial velocity of 50 m s⁻¹ at t = 0 s. At t = 2 s, another ball is projected vertically upward with same velocity. At t = ______ s, second ball will meet the first ball (g = 10 m s⁻²).

Step-by-step solution