Question

A bullet is fired from a gun with a speed of 100m/s in a direction of 30° above the horizontal then calculate maximum height reached time of flight and horizontal range​

Answer 1

A bullet is fired from a gun with a speed of 100m/s in a direction of 30° above the horizontal then calculate maximum height reached time of flight and horizontal range

Answer:

Maximum height = 125 m

Time of flight = 5 seconds

Horizontal range = 250√3 m

Solution:

It is clear that the motion of the bullet can be categorised as a projectile motion. Let's find the required variables from the question,

Initial velocity, u = 100 m/s

Angle with the horizontal, = 30°

We have to find,

Maximum height reached by the bullet.

Time of flight.

Horizontal range.

For maximum height, we have the following formula,

⇒ Hₘₐₓ = u² sin² / 2g

⇒ Hₘₐₓ = (100)² (sin 30°)² / 2×10

⇒ Hₘₐₓ = 10⁴ (1/2)² / 20 [ sin 30° = 1/2 ]

⇒ Hₘₐₓ = 10³ / 2×4

⇒ Hₘₐₓ = 1000 / 8

⇒ Hₘₐₓ = 125 m

Now, Let's find Time of flight, which we is given by the formula,

⇒ T = (Initial velocity in vertical direction) / g

⇒ T = u sin / g

⇒ T = 100×sin 30° / 10 [ sin 30° = 1/2 ]

⇒ T = 10 × 1/2

⇒ T = 5 seconds

Furthermore, For getting the horizontal range, we use the following formula,

⇒ R = u² sin 2 / 2g

⇒ R = (100)² × sin (2×30) / 2×10

⇒ R = 10000 × sin 60° / 20

⇒ R = 500 × √3/2 [ sin 60° = √3/2 ]

⇒ R = 250√3 m

Answer 2

Explanation:

A bullet is fired from a gun with a speed of 100m/s in a direction of 30° above the horizontal then calculate maximum height reached time of flight and horizontal range

Answer:

Maximum height = 125 m

Time of flight = 5 seconds

Horizontal range = 250√3 m

Solution:

It is clear that the motion of the bullet can be categorised as a projectile motion. Let's find the required variables from the question,

Initial velocity, u = 100 m/s

Angle with the horizontal, = 30°

We have to find,

Maximum height reached by the bullet.

Time of flight.

Horizontal range.

For maximum height, we have the following formula,

⇒ Hₘₐₓ = u² sin² / 2g

⇒ Hₘₐₓ = (100)² (sin 30°)² / 2×10

⇒ Hₘₐₓ = 10⁴ (1/2)² / 20 [ sin 30° = 1/2 ]

⇒ Hₘₐₓ = 10³ / 2×4

⇒ Hₘₐₓ = 1000 / 8

⇒ Hₘₐₓ = 125 m

Now, Let's find Time of flight, which we is given by the formula,

⇒ T = (Initial velocity in vertical direction) / g

⇒ T = u sin / g

⇒ T = 100×sin 30° / 10 [ sin 30° = 1/2 ]

⇒ T = 10 × 1/2

⇒ T = 5 seconds

Furthermore, For getting the horizontal range, we use the following formula,

⇒ R = u² sin 2 / 2g

⇒ R = (100)² × sin (2×30) / 2×10

⇒ R = 10000 × sin 60° / 20

⇒ R = 500 × √3/2 [ sin 60° = √3/2 ]

⇒ R = 250√3 m