Question

A bullet is fired from a rifle with an initial velocity of 300 m/s at an angle of 25 degrees from the horizontal.
1. Calculate the horizontal and vertical components of the velocity
2. Calculate the time spent in the air
3. Calculate the maximum height
4. Calculate the horizontal distance

Answer 1

Answer:

1. The horizontal component of the velocity is approximately 271.89 m/s

The vertical component of the velocity is approximately 126.79 m/s

2. The time the bullet spends in the air is approximately 25.85 seconds

3. The maximum height reached by the bullet is approximately 819.35 meters

4. The horizontal distance is approximately 7028.36 meters

Explanation:

The initial velocity of the bullet = 300 m/s

The angle at which the bullet was fired, θ = 25° from the horizontal

1. The horizontal component of the velocity vₓ = The component of the velocity in the x-direction

vₓ = v × cos(θ)

Substituting gives;

vₓ = 300 × cos(25°) ≈ 271.89 m/s

The vertical component of the velocity $u_y$ = The component of the velocity in the y-direction

$v_y$ = v × sin(θ)

Substituting gives;

$u_y$ = 300 × sin(25°) ≈ 126.79 m/s

2. The time the bullet takes to maximum height, t = 1/2 × The time the bullet spends in the air

The following is the relevant kinematic equation for the vertical motion of the bullet;

v = u - g × t

Where;

$v_y$ = The final velocity vertical velocity = (0 at maximum height)

$u_y$ = The initial vertical velocity ≈ 126.79 m/s

g = The acceleration due to gravity = 9.81 m/s²

t = The time in flight of the bullet

Substituting gives;

0 = 126.79 - 9.81 × t

t ≈ 126.79/9.81 ≈ 12.924 s

The time the bullet spends in the air = 2 × The time the bullet takes to maximum height

∴ The time the bullet spends in the air ≈ 2 × 12.924 ≈ 25.85

The time the bullet spends in the air ≈ 25.85 seconds

3. The maximum height reached is given from the following kinematic equation;

$u_y$² = 2 × g × h

Where;

h = The maximum height reached by the bullet

∴ h = $u_y$²/(2 × g)

h = 126.79²/(2 × 9.81) = 819.35

The maximum height reached by the bullet = h = 819.35 meters

4. The horizontal distance travelled by the bullet = Horizontal component of the velocity of the bullet × The time the bullet spends in the air

∴ The horizontal distance travelled by the bullet = 271.89 × 25.85 ≈ 7028.36

The horizontal distance travelled by the bullet ≈ 7028.36 meters.