A submarine launched Ballistic Missile (SLBM) is fired from a distance of 3000 km. If the earth is considered flat and the angle of launch is 45° with horizontal, find the time taken by SLBM to hit the target and the velocity with which the missile is fired. 

(Ans: 5.42 km s⁻¹, 13 min)



Given:

Range = R = 3000 km = 3 ｘ10⁶ m
  
Angle = 𝞱 = 45°

Vale of g = 9.8 m s⁻²


To Find:

Time of flight = t = ?

Initial velocity of Missile = `\v_i` = ?


Solution:

Here we have fined the time initial velocity `\v_i` first so using formula of range of projectile

Using the formula of Range of projectile

R = `\frac {v_i^2 sin 2θ}{g}`

or

`\v_i` = `\sqrt frac {Rg}{sin 2θ}`


by putting the corresponding values

`\v_i` = `\sqrt frac {3 ｘ10⁶ m ｘ 9.8 m s^{-2}}{sin 2x45°}`

`\v_i` = `\sqrt frac {29.4 ｘ10⁶ m^2 s^{-2}}{sin 90°}`

`\v_i` = `\sqrt frac {29.4 ｘ10⁶ m^2 s^{-2}}{1}`

`\v_i` = `\sqrt 29.4 ｘ10⁶ m^2 s^{-2}`

`\v_i` = 5.385 ｘ10³ m s⁻¹ 

or

`\v_i` = 5.4 km s⁻¹ -----------Ans (1).



Now the formula to find the total time of flight of a projectile  is given by; 

T = `\frac {2vi sin θ}{g}`

T = `\frac {(2ｘ5.4 ｘ10³ m s⁻¹) sin 45°}{9.8 m s⁻² }`

T = `\frac {(10.77 m s⁻¹ )(0.707)}{9.8 m s⁻² }`

T = 0.779 ｘ10³ s

or

T = 779  s

T = 12.98  min

T = 13 m ----------------Ans.

Thus the time of flight of the missile is 13 min


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