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A particle is thrown with a velocity 60 m/s at an angle of 30 degree from horizontal. If it passes point A and point B at same height at time 2 s and 4 s then height of the point A and B is-?
Most Upvoted Answer
A particle is thrown with a velocity 60 m/s at an angle of 30 degree f...
**Given Data:**
- Initial velocity of the particle, u = 60 m/s
- Angle with horizontal, θ = 30°
- Time taken to pass point A, t₁ = 2 s
- Time taken to pass point B, t₂ = 4 s

**To Find:**
- Height of point A and point B

**Solution:**

1. **Horizontal and Vertical Components of Velocity:**
The initial velocity can be resolved into horizontal and vertical components using trigonometry.

- Horizontal component: uₓ = u * cos(θ)
- Vertical component: uᵧ = u * sin(θ)

Substituting the given values:
- uₓ = 60 * cos(30°) = 60 * √3/2 = 30√3 m/s
- uᵧ = 60 * sin(30°) = 60 * 1/2 = 30 m/s

2. **Vertical Motion:**
The height of the particle at any given time can be calculated using the equation for vertical motion.
- The equation for vertical displacement is:
h = uᵧ * t + (1/2) * g * t²

- At time t = 2 s:
h₁ = uᵧ * t₁ + (1/2) * g * t₁²
= 30 * 2 + (1/2) * 9.8 * 2²
= 60 + 19.6
= 79.6 m

- At time t = 4 s:
h₂ = uᵧ * t₂ + (1/2) * g * t₂²
= 30 * 4 + (1/2) * 9.8 * 4²
= 120 + 19.6 * 8
= 120 + 156.8
= 276.8 m

3. **Conclusion:**
- The height of point A is 79.6 meters.
- The height of point B is 276.8 meters.
Community Answer
A particle is thrown with a velocity 60 m/s at an angle of 30 degree f...
Y= usintheta t - 1/2 gt^2
Y= (60) (1/2)(2) - (1/2) (10)(4)
Y=40 m
A=B=40m
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