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The initial speed of an arrow shot from ...

The initial speed of an arrow shot from a bow, at an elevation of `30^(@)," is "15 ms^(–1)`. Find its velocity when it hits the ground back

Text Solution

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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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