Question

In the war zone, an army tank A is approaching the
enemy tank B as shown in figure. A shell is fired
from tank A with muzzle velocity v, at an angle 37°
to the horizontal at the instant when tank B is 60 m
away. Tank B which is moving away with velocity
60 ms 1 is hit by shell, then v, is (g = 10 m/s2)​

Answer 1

Answer:

v = 82.57 m/s

Explanation:

Let say after t sec Shell hit the Tank B

Tank B Velocity = 60 m/s

Horizontal Distance Covered by Tank B  = 60* t m

Horizontal Distance to be covered by Shell = 60t  + 60 m

shell is fired from tank A with muzzle velocity v, at an angle 37° to the horizontal

=> Horizontal Speed of Sheel = vCos37 = 0.8v  m/s

Distance Covered in t sec = 0.8v * t  m

0.8v * t = 60t  + 60

Time to peak = t/2

Using v = u + at

=> 0 = vSin37 - g(t/2)

=> (t/2) = 0.6v/g

=> t = 1.2v/g

=> t = 0.12v    (taking g = 10 m/s²)

=> 0.8v * 0.12v = 60*0.12v + 60

Dividing by 0.12

=> 0.8v² = 60v + 500

=>  v² = 75v + 625

=> v² - 75v - 625 = 0

=> v = (75 ± √(75²- 4(1)(-625) ) /2

=> v = (75 ± 90.14)/2

=> v = 82.57 m/s

Answer 2

Answer: the answer is 25(1+√2) m/s

Explanation:

Consider the tank A is at rest, then the tank B speed is +40m/s

Let the time taken to hit the tank B be T

Now, the tank B covered distance in T sec = 40T

Now Total distance from tank A to B is (60 + 40T)m eq1

Now the shell is fired from tank A at a angle of 37° with velocity V⁰, is also have velocity is horizontal and vertical

The horizontal velocity = V⁰cos37 = 4/5V⁰

Now from horizontal velocity distance covered in T time = 4/5V⁰T eq2

From eq 1 and 2

The both distance will be same

By equating them

We will get

4V⁰T - 200T = 60. eq3

Now

The time taken to hit the shell is equal to the time of flight of the shell

T = (2V⁰Sin37)/g

Solving this we will get

T value which is 3V⁰/25

Putting the value of T in eq3

And solving it you will get V⁰

Which is 25(1+√2)m/s