Question

A body is projected with initial velocity of 40 m/s and at an angle of 30°with the horizontal. Calculate (i) time of flight and (i i) horizontal range of projectile.(assume g= 10 m/s²)​

Answer 1

Answer:

Time of Flight will be 4 second.

Horizontal range will be 80√3 metres.

Explanation:

According to the Question

It is given that ,

• Initial velocity ,u = 40m/s
• Angle ,θ = 30°
• Acceleration due to gravity ,g = 10m/s²

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we have to calculate the the time of flight .

So, let's calculate. We will use the formula .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☄️ T = 2uSinθ/g

where,

T denote time of flight

u denote initial velocity

g denote acceleration due to gravity

Substitute the value we get

➻ T = 2*40Sin30°/10

➻ T = 80*½/10

➻ T = 40/10

➻ T = 4s

• Hence, the time of flight will be 4 second .

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Now, calculating the horizontal range of the projectile .

Using Formula.

⠀⠀⠀⠀⠀⠀☄️ Horizontal Range = u²Sin2θ/g

substitute the value we get

➻ Horizontal Range = 40²Sin2*30°/10

➻ Horizontal Range = 1600Sin60°/10

➻ Horizontal Range = 1600√3/2/10

➻ Horizontal Range = 800√3/10

➻ Horizontal Range = 80√3m

• Hence, the horizontal range of the projectile will be 80√3 metres.

Answer 2

Question:

• A body is projected with initial velocity of 40 m/s and at an angle of 30° with the horizontal. Calculate (i) time of flight and (i i) horizontal range of projectile. (Assume g= 10 m/s²)

Answer:

• Time of flight = 4 seconds
• Horizontal range of projectile = 80√3 meters

Explanation:

Given that:

• Initial velocity (u) = 40 m/s
• Angle with horizontal (θ) = 30°
• Acceleration due to gravity (g) = 10 m/s²

To Find:

• Time of flight (T)?
• Horizontal range (R)?

Solution:

❶ Finding time of flight :-

We know that,

$\boxed{\bf{T = \dfrac{2uSin\theta}{g}}}$

Where,

• T is time
• u is initial velocity
• θ is angle with horizontal
• g is acceleration due to gravity

According to the question putting all values in formula we get,

➜ $\sf Time = \dfrac{2\:\times\:40\:\times\:Sin30^{\circ}}{10}$

➜ $\sf Time = \dfrac{\cancel{80}\:\times\:\dfrac{1}{\cancel{2}}}{10}$

➜ $\sf Time = {\cancel{\dfrac{40}{10}}}$

➜ $\bf \red{Time = 4\: s}$

∴ Time of flight is 4 seconds.

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❷ Finding horizontal range :-

We know that,

$\boxed{\bf{R = \dfrac{u^{2}Sin2\theta}{g}}}$

Where,

• R is horizontal range
• u is initial velocity
• θ is angle with horizontal
• g is acceleration due to gravity

According to the question putting all values in formula we get,

➜ $\sf R = \dfrac{(40)^2\:\times\:Sin(2\:\times\:30)^{\circ}}{10}$

➜ $\sf R = \dfrac{1600\:\times\:Sin60^{\circ}}{10}$

➜ $\sf R = \dfrac{\cancel{1600}\:\times\:\dfrac{\sqrt{3}}{\cancel{2}}}{10}$

➜ $\sf R = \dfrac{80\cancel{0}\:\times\:\sqrt{3}}{1\cancel{0}}$

➜ $\sf R = \dfrac{80\sqrt{3}}{1}$

➜ $\bf\purple{ R = 80\sqrt{3}\:m}$

∴ Horizontal range of projectile is 80√3 meters.

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• brainly.in/question/43400404

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