A particle is projected from ground with a speed of v0=80 m/s at angle of 60∘ with horizontal. During first 3 seconds average velocity is →v & acceleration is →a & displacement is −→Δr, then (→a×→v).−→Δr equals to

A particle is projected from ground with a speed of v0=80 m/s at angle

1.	A particle is projected from ground with a speed of v0=80 m/s at angle of 60∘ with horizontal. During first 3 seconds average velocity is →v & acceleration is →a & displacement is −→Δr, then (→a×→v).−→Δr equals to
Answer» A particle is projected from ground with a speed of $v_{0} = 80 m / s$ at angle of $60^{\circ}$ with horizontal. During first 3 seconds average velocity is $\to v$ & acceleration is $\to a$ & displacement is $- \to Δ r$ , then $(\to a \times \to v) . - \to Δ r$ equals to

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A particle is projected from ground with a speed of v0=80 m/s at angle of 60∘ with horizontal. During first 3 seconds average velocity is →v &amp; acceleration is →a &amp; displacement is −→Δr, then (→a×→v).−→Δr equals to

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A particle is projected from ground with a speed of v0=80 m/s at angle of 60∘ with horizontal. During first 3 seconds average velocity is →v & acceleration is →a & displacement is −→Δr, then (→a×→v).−→Δr equals to