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| 1. |
Prove that the maximum horizontal range 'is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal |
| Answer» Solution :`R_max.=mu^2/g,` when `theta=45^@` Then the height ATTAINED `H=(mu^2(SIN45^@)^2)/(2G)=mu^2/(4g)=(R_max)/4` THUS `R_max=4H` | |