In a wedding, a bullet is fired in the air at a speed of 500 m/s making an angle of 60° horizontal from an AK 47 rifle. (a) How high will the bullet rise? (b) What time would it take to reach the ground? (c) How far would it go? (Ignore air resistance)
((a) 9,560 m (b) 88.3 s (c)22,078 m)


Given:

Initial velocity of the bullet = vi = 500 m s⁻¹  
Angle = 𝞱 = 60°
Vale of g = 9.81 m s⁻²

To Find:

(a) Maximum height of bullet rise = H = ?
(b) Time of flight of bullet = T = ?
(c) Range of bullet = R = ?



Solution:

(a) Height of bullet rise = H = ?

As the bullet fired from the gun will follow the path of  projectile motion so, the formula to find the height reached by a projectile is given; 


H = `\frac {vi² sin² θ}{2g}`

H = `\frac {(500 m s⁻¹ )²  sin² 60}{2ｘ9.81 m s⁻² }`

H = `\frac {(250000 m² s⁻² )(0.75)}{19.62 m s⁻² }`

H = 9556.975 m

or

H = 9567 m ----------------Ans.

Thus the maximum height reached by the bullet is 9566 m



(b) Time of flight of the bullet = T = ?


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

T = `\frac {2vi sin θ}{g}`

T = `\frac {(2ｘ500 m s⁻¹ ) sin 60}{9.81 m s⁻² }`

T = `\frac {(1000 m s⁻¹ )(0.855)}{9.81 m s⁻² }`

T = 88.360 s

or

T = 88.3 s ----------------Ans.

Thus the time of flight of the bullet is 88.4 s


(c) Range of bullet = R = ?

The formula to find the Range of a projectile is given; 

R = `\frac {vi² sin (2θ)}{g}`

R = `\frac {(500 m s⁻¹ )² sin (2ｘ60°)}{9.81 m s⁻² }`

R = `\frac {(250000 m² s⁻² ) sin (120°)}{9.81 m s⁻² }`

R = `\frac {(250000 m² s⁻² )(0.866)}{9.81 m s⁻² }`

R = 22069.317 m

or

R = 22069 m ----------------Ans.

Thus the Range of the bullet is 22069 m




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