Physics, asked by Anonymous, 1 year ago

\displaystyle \text{Give answer if you know only}\\\\\\\displaystyle \text{The range of projectile in meters (round off to closest integer) on the }\\\\\\\displaystyle \text{ inclined plane which is projected perpendicular to the incline plane }\\\\\\\displaystyle \text{with velocity 20 m/s as shown in figure is $(70+x)(Take \ g=10 m/s^2$}\\\\\\\displaystyle \text{and $\sin37=0.6$)}

\displaystyle \text{Options are :}\\\\\\\displaystyle \text{a. \ 85 \ m}\\\\\\\displaystyle \text{b. \ 65 \ m}\\\\\\\displaystyle \text{ c. \ 75 \ m}\\\\\\\displaystyle \text{d. \ 80 \ m}\\\\\\\displaystyle \text{Kindly give good and contain quality answer with great explain.}

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Answered by Anonymous
7

what is x here?

well range is 75 So x should be 5

But according to options I think its range So its 75

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Answered by mustafa3952
0

Answer:

c.75 m

Explanation:

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