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A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ?
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A ball is projected to an angle 60 with horizontal at initial velocity...
The speed of a projectile can be calculated using the principles of projectile motion. In this case, we are given that a ball is projected at an angle of 60 degrees with the horizontal and an initial velocity of 20 m/s. We need to find the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal.
To solve this problem, we can break it down into two components: the horizontal and vertical components of the velocity.
1. Horizontal Component:
The horizontal component of the velocity remains constant throughout the motion. It can be calculated using the initial velocity and the angle of projection. In this case, the angle of projection is 60 degrees.
Horizontal component of velocity (Vx) = V * cos(theta)
Vx = 20 m/s * cos(60)
Vx = 20 m/s * 0.5
Vx = 10 m/s
2. Vertical Component:
The vertical component of the velocity changes due to the effect of gravity. It can be calculated using the initial velocity and the angle of projection. In this case, the angle of projection is 60 degrees.
Vertical component of velocity (Vy) = V * sin(theta)
Vy = 20 m/s * sin(60)
Vy = 20 m/s * 0.866
Vy = 17.32 m/s
3. Finding the Speed:
To find the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal, we need to calculate the resultant velocity (Vr) using the horizontal and vertical components.
Vr = sqrt(Vx^2 + Vy^2)
Vr = sqrt((10 m/s)^2 + (17.32 m/s)^2)
Vr = sqrt(100 m^2/s^2 + 300 m^2/s^2)
Vr = sqrt(400 m^2/s^2)
Vr = 20 m/s
Therefore, the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal is approximately 20 m/s.
Summary:
- The horizontal component of velocity remains constant throughout the motion and is calculated using V * cos(theta).
- The vertical component of velocity changes due to the effect of gravity and is calculated using V * sin(theta).
- The speed of the particle is found by calculating the resultant velocity using the horizontal and vertical components.
- In this case, the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal is approximately 20 m/s.
To solve this problem, we can break it down into two components: the horizontal and vertical components of the velocity.
1. Horizontal Component:
The horizontal component of the velocity remains constant throughout the motion. It can be calculated using the initial velocity and the angle of projection. In this case, the angle of projection is 60 degrees.
Horizontal component of velocity (Vx) = V * cos(theta)
Vx = 20 m/s * cos(60)
Vx = 20 m/s * 0.5
Vx = 10 m/s
2. Vertical Component:
The vertical component of the velocity changes due to the effect of gravity. It can be calculated using the initial velocity and the angle of projection. In this case, the angle of projection is 60 degrees.
Vertical component of velocity (Vy) = V * sin(theta)
Vy = 20 m/s * sin(60)
Vy = 20 m/s * 0.866
Vy = 17.32 m/s
3. Finding the Speed:
To find the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal, we need to calculate the resultant velocity (Vr) using the horizontal and vertical components.
Vr = sqrt(Vx^2 + Vy^2)
Vr = sqrt((10 m/s)^2 + (17.32 m/s)^2)
Vr = sqrt(100 m^2/s^2 + 300 m^2/s^2)
Vr = sqrt(400 m^2/s^2)
Vr = 20 m/s
Therefore, the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal is approximately 20 m/s.
Summary:
- The horizontal component of velocity remains constant throughout the motion and is calculated using V * cos(theta).
- The vertical component of velocity changes due to the effect of gravity and is calculated using V * sin(theta).
- The speed of the particle is found by calculating the resultant velocity using the horizontal and vertical components.
- In this case, the speed of the particle when the velocity makes an angle of 45 degrees with the horizontal is approximately 20 m/s.
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A ball is projected to an angle 60 with horizontal at initial velocity...
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A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ?
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A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ?.
A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is projected to an angle 60 with horizontal at initial velocity 20m/s. The speed of particle when velocity makes an angle 45 with horizontal is nearly ?.
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