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A particle is projected from ground with velocity 50 m/s at 37 degrees find velocity and displacement after 2 seconds?
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A particle is projected from ground with velocity 50 m/s at 37 degrees...
Given:
Initial velocity (u) = 50 m/s
Launch angle (θ) = 37 degrees
Time (t) = 2 seconds

Calculating Vertical Velocity (Vy):
Using the formula for vertical velocity, Vy = u * sin(θ), we can calculate the vertical component of the velocity.

Vy = 50 * sin(37°)
Vy ≈ 30.1 m/s

Calculating Horizontal Velocity (Vx):
Using the formula for horizontal velocity, Vx = u * cos(θ), we can calculate the horizontal component of the velocity.

Vx = 50 * cos(37°)
Vx ≈ 40.0 m/s

Calculating Displacement in the Vertical direction (Sy):
Using the formula for displacement in the vertical direction, Sy = u * t * sin(θ) - (1/2) * g * t^2, we can calculate the displacement.

Sy = 50 * 2 * sin(37°) - (1/2) * 9.8 * (2^2)
Sy ≈ 30.1 * 2 * 0.6018 - 19.6
Sy ≈ 35.24 - 19.6
Sy ≈ 15.64 m

Calculating Displacement in the Horizontal direction (Sx):
Using the formula for displacement in the horizontal direction, Sx = u * t * cos(θ), we can calculate the displacement.

Sx = 50 * 2 * cos(37°)
Sx ≈ 40.0 * 2 * 0.7986
Sx ≈ 63.82 m

Calculating Total Displacement (S):
The total displacement can be calculated using the Pythagorean theorem.

S = sqrt(Sx^2 + Sy^2)
S = sqrt(63.82^2 + 15.64^2)
S ≈ sqrt(4072.9924 + 244.3696)
S ≈ sqrt(4317.362)
S ≈ 65.73 m

Final Results:
- Vertical Velocity (Vy) ≈ 30.1 m/s
- Horizontal Velocity (Vx) ≈ 40.0 m/s
- Displacement in the Vertical direction (Sy) ≈ 15.64 m
- Displacement in the Horizontal direction (Sx) ≈ 63.82 m
- Total Displacement (S) ≈ 65.73 m
Community Answer
A particle is projected from ground with velocity 50 m/s at 37 degrees...
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A particle is projected from ground with velocity 50 m/s at 37 degrees find velocity and displacement after 2 seconds?
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