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Points A and B are market on the ground at a distance 1 m from each other . C is marked as midpoint between A and B . O is point on the same line at a distance 5. m from point C. A particle is projected with a speed 10 m//s from point O. What should ne the angle of projection so that particfle strikes somewhere between A and B ? |
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Answer» Solution :Rangeof the projectile can be written as follow: `R=(u^(2) sin 2 thetqa)/(g) =(10 sin 2theta)/(10)` `R=1- sin 2 theta` Range of the projectile can be plotted against `theta` as shown in the figure .  For the particle to land between A and B , its range must be between 5 m and 6m Let `alpha ` be the angle of projection corresponding to R=5m .`R=10 sinsin 2 aplha =5 rArr sin 2 alpha =1//2 rArr 2alpha =30^(@) rArr alpha 15^(@)` Range will be samefor `(90-alpha)=75^(@)` Similarly we can CALCULATE angleof projection `beta` for R=6 m . `R=10 sin 2 beta =6 rArr sin 2 beta =3//5 rArr 2 beta =37^(@) rArr beta =18.5^(@)` Rnage will be same for `(90-beta)=71.5^(@)` We can understand that range is between 5m and6m when angle of projection is between `15^(@)`and `18.5^(@)` and alsowhen angle of projection is between `71.5^(@)` and `75^(@)` . From the graph we can understand thatrange of projectile is GREATER than 6 m when angle of projectionis between `18.5^(@)` and `71.5^(@)` andthisinterval is `(71.5-18.5)=53^(@)` . Hence we can also SAY thatminimum angle of projection is `15^(@)` and maximumis `75^(@)` which is not allowed for correct landing of particle. |
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