Question

5. A body is projected with a velocity of 10 m/s at 45° to the horizontal. The velocity of the projectile when it moves at 30° to this horizontal is
$your answer \: has \: to \: come \sqrt{200 \div 3} m \: per \: second$
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Answer 1

Given data : body is projected with a velocity of 10 m/s at 45° to the horizontal.

To find : The velocity of the projectile when it moves at 30° to this horizontal is ?

Solution : a/c to question, body is projected with a velocity of 10 m/s at 45° to the horizontal.

Let, initial velocity of body, u = 10 m/s. a/c to given data we take horizontal componet, cos θ = 45⁰.

simllarly,

When the body attains its final velocity v, it is at an angle of 30⁰ to the horizontal. so we take horizontal component cos θ = 30⁰.

Law of projectile motion : The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity.The only force acting upon a projectile is gravity.

Hence, a/c to figure and law.

u cos θ = v cos θ

Now,

⟹ u cos θ = v cos θ

⟹ 10 * cos 45⁰ = v cos 30⁰

⟹ 10 * 1/√2 = v * √3/2

⟹ v * √3/2 = 10/√2

⟹ v * √3 = 10/√2 * 2

⟹ v * √3 = 200 /√2

⟹ v * √3 = √(400/2)

⟹ v = √(400/2)/√3

⟹ v = √(400/6)

⟹ v = √200/3 m/s

Answer : Hence, the velocity of the projectile when it moves at 30° to this horizontal is √200/3 m/s.

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Answer 2

• 10 cos 45= v cos 30
• Please understand in the given picture