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A particle is ejected from the tube at A with a velocity `V` at an angle `theta` with the verticle y-axis at a height h above the ground as shown. A strong horizontal wind gives the particle a constant horizontal acceleration a in the positive x-direction. If the particle strikes the ground at a point directly under its released position and the downward y-acceleration is taken as g then find h. A. `h= (2V^(2)sin theta cos theta)/(a)`B. `h =(2V^(2)sin theta cos theta)/(g)`C. `h =(2V^(2))/(g)sin theta[cos theta+(a)/(g)sin theta]`D. `h =(2V^(2))/(a)sin theta[cos theta+(g)/(a)sin theta]` |
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Answer» Correct Answer - D A particle is ………… Since x - displecement is zero `0 = Vsin theta t-(1)/(2)at^(2)` `t = (2Vsin theta)/(a)` `h = Vcos theta t+(1)/(2)"gt"^(2)` `h = (2V^(2))/(a)sin theta [cos theta+(g)/(a)sin theta]` |
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