answer the following questions correct plzzzzzzzzz see image questions full
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Answer : R
t1=2vsinθgt1=2vsinθg
t2=2vsin(90∘−θ)g=2vcosθgt2=2vsin(90∘−θ)g=2vcosθg
Now,t1t2=(2vsinθ)(2vcosθ)g2t1t2=(2vsinθ)(2vcosθ)g2
⇒2g[v2(2sinθcosθ)g]=2gv2sin2θg=2g=⇒2g[v2(2sinθcosθ)g]=2gv2sin2θg=2g=RR
∴t1t2∝R
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t1=2vsinθgt1=2vsinθg
t2=2vsin(90∘−θ)g=2vcosθgt2=2vsin(90∘−θ)g=2vcosθg
Now,t1t2=(2vsinθ)(2vcosθ)g2t1t2=(2vsinθ)(2vcosθ)g2
⇒2g[v2(2sinθcosθ)g]=2gv2sin2θg=2g=⇒2g[v2(2sinθcosθ)g]=2gv2sin2θg=2g=RR
∴t1t2∝R
hope this help ful for you.
mark as brainliest.
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