time displacement height

Question

1.3 A bullet is fired from a rifle at an angle of 20° to the horizontal with an initial velocity of 250 m/s. Calculate the following: 1.3.1 The time for the bullet to reach the maximum height 1.3.2 The maximum height reached by the bullet 1.3.3 The horizontal displacement of the bullet

Answer 1

1.3.1)

Given data :

Fired angle θ = 20°

Initial velocity v₀ =  250 m/s

Earth gravity g = 9.8 m/s

The time for the bullet to reach the maximum height t_max = ?

Formula :

Maximum time t_max = (v₀sinθ)/g

Substitute : v₀ =  250 m/s , θ = 20° and  g = 9.8 m/s.

t_max = (250sin20) / (9.8)

t_max = (250×0.342) / (9.8)

t_max = 8.72 sec

The time for the bullet to reach the maximum height :  t_max = 8.72 sec

 

Answer 2

1.3.2)

Given data :

Fired angle θ = 20°

Initial velocity v₀ =  250 m/s

Earth gravity g = 9.8 m/s

The maximum height reached by the bullet H_max = ?

Formula :

Maximum height H_max = (v₀²sin²θ)/2g

Substitute : v₀ =  250 m/s , θ = 20° and  g = 9.8 m/s.

H_max = (250²sin²20) / (2×9.8)

H_max = (62500×0.0119) / (19.6)

H_max = 373.02 m

Maximum Height  H_max = 373.02 m.

Answer 3

1.3.3)

Given data :

Fired angle θ = 20°

Initial velocity v₀ =  250 m/s

Earth gravity g = 9.8 m/s

The horizontal displacement reached by the bullet d = ?

Formula :

Horizontal displacementd = (v₀²sin2θ)/g

Substitute : v₀ =  250 m/s , θ = 20° and  g = 9.8 m/s.

d = (250²sin(2×20)) / (9.8)

d = (250²(0.64)) / (9.8)

d = 4099.41 m

The horizontal displacement is 4099.41 meters