Question

Two bodies are projected vertically upwards from one point with the same initial velocities v0 m/s. The second body is thrown τ s after the first. The two bodies meet after time

- v0g−τ2
- v0g+τ
- v0g+τ2
- v02g−τ

Solution

The correct option is **C** v0g+τ2

Height of first body after time t,h1=v0t−12gt2

Height of the second body after time (t−τ),

h2=v0(t−τ)−12g(t−τ)2

If they meet after time t, h1=h2

⇒t=v0g+τ2

The correct option is (c).

Height of first body after time t,h1=v0t−12gt2

Height of the second body after time (t−τ),

h2=v0(t−τ)−12g(t−τ)2

If they meet after time t, h1=h2

⇒t=v0g+τ2

The correct option is (c).

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