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

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

Answer:

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Given :-

A bullet is fired from a gun with a speed of 100 m/s in a direction of 30° above the horizontal.

To Find :-

(i) Maximum Height

(ii) Time Taken

(iii) Horizontal Range

Formula Used :-

➲ Maximum Height :-

⇒ MaximumHeight= 2gv²sin²θ

➲ Time Taken :

⇒ TimeTaken= g²vsinθ

➲ Horizontal Range :

⇒ Horizontal Range= gv²sin2θ

where,

v = Velocity

g = Acceleration due to gravity

Solution :-

➲ To find maximum height :

Given :

Velocity (v) = 100 m/s

θ = 30°

Acceleration due to gravity (g) = 10 m/s²

According to the question by using the formula we get,

⇒ MaximumHeight = 2×10(100)² × (sin30°)²

As we know that (sin30° = 1/2),

⇒ MaximumHeight = 20100×100×(21)²

⇒ MaximumHeight = 2010000 × 41

⇒ MaximumHeight = 202500 × 1

⇒ MaximumHeight = 202500

⇒ Maximum Height = 2250

⇒ Maximum Height = 125m

∴ The maximum height is 125 m .

➲ To find time taken :

Given :

Velocity (v) = 100 m/s

θ = 30°

Acceleration due to gravity (g) = 10 m/s²

According to the question by using the formula we get,

⇒ Time Taken = 10²×100×sin30°

As we know that, (sin30° = 1/2)

⇒ Time Taken = 10200 × 21

⇒ Time Taken = 10100

⇒ Time Taken = 10seconds

∴ The time taken to flight is 10 seconds

➲ To find horizontal range :

Given :

Velocity (v) = 100 m/s

θ = 30°

Acceleration due to gravity (g) = 10 m/s²

According to the question by using the formula we get,

⇒ Horizontal Range = 10(100)²×sin2×30°

⇒ Horizontal Range= 1010000×sin60°

As we know that, (sin60° = √3/2)

⇒ Horizontal Range= 1010000 × 23

⇒ Horizontal Range= 5000 × 3

⇒ Horizontal Range=500× 3

⇒ Horizontal Range = 500³

∴ The horizontal range is 500√3.

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