InterviewSolution
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InterviewSolution
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A gun is fired from a moving platform and the ranges of theshot are observed to be R and S when platform is moving forward or backward respectively with velocity v prove that the elevation of the gun is `tan ^(-1)[(g(R-S)^(2))/(4V^(2)(R+S))]` |
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Answer» Let u and v be the horizontal and vertical components of velocity when the platform is at rest. When the platform is moving forward, the rsu ltant horizontal velocity is `=u+V` and vertical velocity is v. the range R is given by `R=(2(u+V)v)/(g)`.....(i) When the plat form is moving backward with velocity V, the horizontal and vertical components of the velocity of the shot are u-V and v respectively. The range is then `S=(2(u-V)v)/(g)` .......(ii) From Eqns. (i) and (ii), `R+S=4u v//g` and `R-S=4Vv//g` `therefore ((R-S)^(2))/((R+S))=(16J//^(2)v^(2)//g^(2))/(4uv//g)=(4v^(2)v)/(ug)` If `alpha` be the angle of projection, then `tan alpha=v//u` Thus, `((R-S)^(2))/((R+S))=(4V^(2))/(g) tan alpha` `tan alpha=(g(R-S)^(2))/((R+S))` or `alpha=tan^(-1)[(g(R-S)^(2))/(4V^(2)(R+S))]` |
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