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At t = 0 a projectile is fired from a point O(taken as origin) on the ground with a speed of `50 m//s` at an angle of `53^(@)` with the horizontal. It just passes two points A & B each at height 75 m above horizontal as shown The horizontal separation between the points A and B isA. 30 mB. 60 mC. 90 mD. None |
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Answer» Correct Answer - D `v_(1)sintheta_(1)=v_(2)sintheta_(2)` Time of flight `T=(2xsintheta)/(g)` `T_(1)=(2v_(1)sintheta_(1))/(g)=(2v_(2)sintheta_(2))/(g)=T_(2)` Maximum height `H_("max")=(x^(2)sin^(2)theta)/(2g)=((xsintheta))/(2g)` `H_("max")=((v_(1)sintheta_(1))^(2))/(2g)=((v_(2)sintheta_(2)))/(2g)=H_("2max")` `vec(a)_("rel")=-ghat(j)+ghat(j)=0` `because vec(a)_("rel")=0` & time of flight is same the trajectory of one w.r.t. other is a straight line. |
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