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A shell fired from a gun at sea level rises to a maximum height of 55 km when fired at a ship 2020 km away. The muzzle velocity should be |
| Answer» VELOCITY of shell is 306.32 km/hr•Muzzle velocity is the velocity with which bullet is fired from gun let the initial velocity of gun be uGiven,•Range of shell = 2020 Km Also , range of projectile is R = (u²sin2A)/g 2020 = (u²sin2A)/g _____(1)•here A is angle at which shell is fired.•Now , maximum height attained by shell is 55 km Also, Maximum height of projectile is :H = (u²sin²A)/2g55 =( u²sin²A)/2G ________(2)•Dividing (1)&(2)2020/55 = {(u²sin2A)/g}/{( u²sin²A)/2g}2020/55 = 2sin2A/sin²A2020/55 = 2(2sinAcosA)/sin²A2020/55 = 4sinAcosA/sin²A2020/220 = cosA/sinA101/11 = cosA/sinAsquaring both sides10201/121 = cos²A/sin²A10201/121 = 1-sin²A/sin²A10201sin²A = 121 -121sin²A(10201+121)sin²A = 12110322sin²A = 121 sin²A = 121/10322 _____(3)•Now, putting (3) in (2)55 = [u² (121/10322)]/2g55×10322 = (u²×121)/2(10)(55×10322×20)/121 = u²u² = 93836.36u = 306.32 Km/sec•Muzzle velocity of shell is 306.32 km/hr | |