Question

A bullet is fired at an angle of 60 degrees with an initial velocity of 200.0 m/s. How long is the bullet in the air? What is the maximum height reached by the bullet?

Answer 1

Answer:

     Time of flight,  t =   35.34 sec.

Maximum heght, H  = 1530.60 m.

Explanation:

initial velocity of the bullet u = 200m/s

Angle of projection  θ =  60°

(a).

As we know that in projectile motion;

Time of flight t = [2u*sin( θ)]/ g

                      t = 2*200*sin ( 60° )/9.8

                    t   = 35.34 sec.

(b).

As we khow that in projectile motion;

Maximum heght, H = [u²*sin²θ]/2g

                                 = (200) ²* sin² (60) /2*9.8

Maximum heght, H  = 1530.60 m.