InterviewSolution
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InterviewSolution
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The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomen such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing. A tower is at a distance of `5m` from a man who can throw a stone with a maximum speed of `10m//s`. What is the maximum height that the man can hit on this tower.A. `4.5m`B. `4m`C. `5m`D. `3.75m` |
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Answer» `tantheta=(u^(2))/(Rg)` `H=R tantheta-(1)/(2)g.(R^(2))/(u^(2))(1+tan^(2)theta)` `=(u^(2))/(g)-(1)/(2)g. (R^(2))/(u^(2))-(1)/(2)(u^(2))/(g)=10-1.25-5=3.75m` |
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