If a projectile is fired with an initial velocity of V meters per second at an angle A above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is given by the parametric equations

x=(Vcos(A))t,
y=(Vsin(A))t−12gt2,

A gun is fired with a = 30° and v0 = 500 m/s. When will the bullet hit the ground?

The projectile will hit the ground after about 1min 41 seconds

T=100.7 seconds

Explanation:

The time of flight for a projectile motion is expressed as

T=(2usin0)/g

Where u is the initial velocity

Theta 0 is the projected angle

Assuming g=9.81m/s

Given data

Theta 0= 30°

u =500m/s

Substituting our data into the expression we have

T=(2*500*sin30°)/9.81

T=(1000*−0.988)/9.81

T=988/9.81

T=100.7 second

The projectile will hit the ground after about 1min 41 seconds

this would be about 43%