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
Answers
Muzzle 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 u
Given,
•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)/2g
55 =( u²sin²A)/2g ________(2)
•Dividing (1)&(2)
2020/55 =
{(u²sin2A)/g}/{( u²sin²A)/2g}
2020/55 = 2sin2A/sin²A
2020/55 = 2(2sinAcosA)/sin²A
2020/55 = 4sinAcosA/sin²A
2020/220 = cosA/sinA
101/11 = cosA/sinA
squaring both sides
10201/121 = cos²A/sin²A
10201/121 = 1-sin²A/sin²A
10201sin²A = 121 -121sin²A
(10201+121)sin²A = 121
10322sin²A = 121
sin²A = 121/10322 _____(3)
•Now, putting (3) in (2)
55 = [u² (121/10322)]/2g
55×10322 = (u²×121)/2(10)
(55×10322×20)/121 = u²
u² = 93836.36
u = 306.32 Km/sec
•Muzzle velocity of shell is 306.32 km/hr