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two particles are projected from a towe horizontally in opposite direction with velocities 20m/s and 10m/s.find the time when thier velocity vectors are mutually perpendicular.take g=10m/s^2.?
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?two particles are projected from a towe horizontally in opposite dire...
Given:
- Initial velocities of the particles: 20 m/s and 10 m/s
- Acceleration due to gravity: 10 m/s^2
To find:
The time at which the velocity vectors of the particles are mutually perpendicular.
Explanation:
When two vectors are mutually perpendicular, their dot product is zero.
Let's assume the time at which the velocity vectors are mutually perpendicular is t seconds.
Step 1: Analyzing the motion of the particles
When a particle is projected horizontally, its initial vertical velocity is zero. Therefore, the vertical motion of the particles can be analyzed separately.
Particle 1:
- Initial horizontal velocity (ux1): 20 m/s
- Vertical acceleration (ay1): -10 m/s^2 (negative because the particle is moving upwards against gravity)
Particle 2:
- Initial horizontal velocity (ux2): -10 m/s (opposite direction to particle 1)
- Vertical acceleration (ay2): -10 m/s^2
Step 2: Finding the vertical displacements of the particles
Using the equation of motion, the vertical displacement (sy) of a particle can be calculated as:
sy = uy * t + (1/2) * a * t^2
where uy is the initial vertical velocity, a is the vertical acceleration, and t is the time.
Particle 1:
sy1 = 0 * t + (1/2) * (-10) * t^2
sy1 = -5t^2
Particle 2:
sy2 = 0 * t + (1/2) * (-10) * t^2
sy2 = -5t^2
Step 3: Finding the horizontal displacements of the particles
Since the particles are projected horizontally, their horizontal displacements (sx) can be calculated using the equation:
sx = ux * t
where ux is the initial horizontal velocity and t is the time.
Particle 1:
sx1 = 20t
Particle 2:
sx2 = -10t
Step 4: Finding the dot product of the velocity vectors
The dot product of two vectors is given by the equation:
A · B = Ax * Bx + Ay * By
where Ax and Ay are the x and y components of vector A, and Bx and By are the x and y components of vector B.
The velocity vectors of the particles are given by:
Particle 1: V1 = 20i - 10jt
Particle 2: V2 = -10i - 10jt
The dot product of these vectors is:
V1 · V2 = (20)(-10) + (-10)(-10)
V1 · V2 = -200 + 100
V1 · V2 = -100
Step 5: Setting the dot product equal to zero
Since the velocity vectors are mutually perpendicular, their dot product is zero. Therefore, we can equate the dot product to zero and solve for t.
-100 = 0
This equation has no solution, which means the velocity vectors of the particles are never mutually perpendicular.
Conclusion:
- Initial velocities of the particles: 20 m/s and 10 m/s
- Acceleration due to gravity: 10 m/s^2
To find:
The time at which the velocity vectors of the particles are mutually perpendicular.
Explanation:
When two vectors are mutually perpendicular, their dot product is zero.
Let's assume the time at which the velocity vectors are mutually perpendicular is t seconds.
Step 1: Analyzing the motion of the particles
When a particle is projected horizontally, its initial vertical velocity is zero. Therefore, the vertical motion of the particles can be analyzed separately.
Particle 1:
- Initial horizontal velocity (ux1): 20 m/s
- Vertical acceleration (ay1): -10 m/s^2 (negative because the particle is moving upwards against gravity)
Particle 2:
- Initial horizontal velocity (ux2): -10 m/s (opposite direction to particle 1)
- Vertical acceleration (ay2): -10 m/s^2
Step 2: Finding the vertical displacements of the particles
Using the equation of motion, the vertical displacement (sy) of a particle can be calculated as:
sy = uy * t + (1/2) * a * t^2
where uy is the initial vertical velocity, a is the vertical acceleration, and t is the time.
Particle 1:
sy1 = 0 * t + (1/2) * (-10) * t^2
sy1 = -5t^2
Particle 2:
sy2 = 0 * t + (1/2) * (-10) * t^2
sy2 = -5t^2
Step 3: Finding the horizontal displacements of the particles
Since the particles are projected horizontally, their horizontal displacements (sx) can be calculated using the equation:
sx = ux * t
where ux is the initial horizontal velocity and t is the time.
Particle 1:
sx1 = 20t
Particle 2:
sx2 = -10t
Step 4: Finding the dot product of the velocity vectors
The dot product of two vectors is given by the equation:
A · B = Ax * Bx + Ay * By
where Ax and Ay are the x and y components of vector A, and Bx and By are the x and y components of vector B.
The velocity vectors of the particles are given by:
Particle 1: V1 = 20i - 10jt
Particle 2: V2 = -10i - 10jt
The dot product of these vectors is:
V1 · V2 = (20)(-10) + (-10)(-10)
V1 · V2 = -200 + 100
V1 · V2 = -100
Step 5: Setting the dot product equal to zero
Since the velocity vectors are mutually perpendicular, their dot product is zero. Therefore, we can equate the dot product to zero and solve for t.
-100 = 0
This equation has no solution, which means the velocity vectors of the particles are never mutually perpendicular.
Conclusion:
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?two particles are projected from a towe horizontally in opposite direction with velocities 20m/s and 10m/s.find the time when thier velocity vectors are mutually perpendicular.take g=10m/s^2.?
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