Physics, asked by anushree11deshmikh, 1 year ago

answer the following questions correct plzzzzzzzzz see image questions full

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Answered by kavita9612
1
Answer : R

t1=2vsinθgt1=2vsin⁡θg

t2=2vsin(90∘−θ)g=2vcosθgt2=2vsin⁡(90∘−θ)g=2vcos⁡θg

Now,t1t2=(2vsinθ)(2vcosθ)g2t1t2=(2vsin⁡θ)(2vcos⁡θ)g2

⇒2g[v2(2sinθcosθ)g]=2gv2sin2θg=2g=⇒2g[v2(2sin⁡θcos⁡θ)g]=2gv2sin⁡2θg=2g=RR

∴t1t2∝R


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