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A bullet is fired at an angle of 15^(@) with the horizontal and it hits the ground 3 km away. Can we hit a target at a distance of 7 km by just adjusting its angle of projection ? |
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Answer» Solution :Range of a projectile, `R=(u^(2) sin 20)/(G)` Here, `R=3 km =3000 m" when "0=15^(@)` `? 3000 =(u^(2) sin 30^(@))/(g) =(u^(2))/(g) xx (1)/(2)` or `(u^(2))/(g)=6000 m =6 km` ? MAXIMUM range `=(u^(2))/(g) =6 km` Hence the bullet can neverhitthe target, whichis 7 km, because its maximum range is 6 km. |
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