A bullet is shot straight up with an initial speed of 180 m/s. How long is the bullet in the air and how high does the bullet go? (neglect air resistance)

Explanation:

Time of flight is the time taken by the bulet in air. Time of flight is expressed as T = Usinθ/g where;

U is the initial velocity = 180m/s

θ is the angle of launch = 90° (Since the bullet is shot upwards)

g is the acceleration due to gravity = 9.81m/s²

T = 180sin90/9.81

T = 180/9.81

T = 18.35s

Hence the bullet will spend 18.35s in air

The height covered by the bullet is the maximum height. Maximum height is expressed as;

H = u²sin²θ/2g

H = 180²sin²90/2(9.81)

H = 180/19.62

H = 9.17m

Hence the bullet will go 9.17m high

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

answergravity;