InterviewSolution
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InterviewSolution
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The initital speed of an arrow shot from s bow, at an elevation of `30^(@)` , is ` 15 m s^(-1)` , Find the velocity when it hits the ground back. |
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Answer» Here ` v_(0) 15 " m s"^(-1)`, angle of projection , ` theta_(0) = 30^(@)` Therefore, ` v_(0x) = v_(0_ cos theta_(0)` ` = 15 cos(30 ^(@))` ` = ( 15 sqrt3)/2 " m s"^(-1)` And ` v_(0y) = v_(0)sin theta_(0)` ` = 15/2 m s ^(-1)` Horizontal component of velocity remians constant throughout the fight i.e. ` v_(x) = v_(0x) = (15 sqrt3)/2 ms^(-1)` Vertical component of velocity is given by ` v_(y) = v_(0) sin theta_(0) - gt ` Put ` t= T_(f) = ( 2v_(0) sin theta_(0))/g ` ` Rightarrow v_(y) = v_(0) sin theta_(0) -g xx ( 2v_(0) sin theta_(0))/ g` ` v_(y) = - v_(0) sintheta_(0)` `1 v_(y) - 15/2 m s^(-1)` Thus, total velocity ` v = sqrt(v_(x)^(2) + v_(y)^(2))` ` sqrt( ( 15^(2)xx 3)/4 + 15^(2)/4))` `v = 15 m s ^(-1)` Let the final velocity make and angle ` theta` with the positive x - axis , then ` tetha = tan^(-1) (v_(y)/v_(x))` ` theta = tan^(-1) (((-15)/2)/(15 sqrt3/2))` ` = tan^(-1) ((-1)/sqrt3)` ` theta = - 30^(@)` |
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