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?

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.