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 = vi = ?


Solution:

Here we have fined the time initial velocity vi first so using formula of range of projectile

Using the formula of Range of projectile

R = v2isin2θg

or

vi = √Rgsin2θ


by putting the corresponding values

vi = √3ｘ10⁶mｘ9.8ms-2sin2x45°

vi = √29.4ｘ10⁶m2s-2sin90°

vi = √29.4ｘ10⁶m2s-21

vi = √29.4ｘ10⁶m2s-2

vi = 5.385 ｘ10³ m s⁻¹ 

or

vi = 5.4 km s⁻¹ -----------Ans (1).



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

T = 2visinθg

T = (2ｘ5.4ｘ10³ms⁻¹)sin45°9.8ms⁻²

T = (10.77ms⁻¹)(0.707)9.8ms⁻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


************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.