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a particle is projected at some angle with velocity 50 metre per second crosses 20m high wall after 4 seconds from the time of projection. find the angle of projection of the particle
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a particle is projected at some angle with velocity 50 metre per secon...
**Given Information:**

- Initial velocity of the particle, u = 50 m/s
- Height of the wall, h = 20 m
- Time taken to cross the wall, t = 4 seconds

**To find:**

The angle of projection of the particle.

**Assumptions:**

- We will consider the motion of the particle in a vertical plane only, neglecting the effects of air resistance.
- The acceleration due to gravity, g = 9.8 m/s², is constant throughout the motion.
- The particle is projected from the ground level.

**Approach:**

We can solve this problem using the equations of motion in two dimensions.

1. Let's break the initial velocity u into its horizontal and vertical components.
- The horizontal component, uₓ, remains constant throughout the motion, as there is no horizontal acceleration. Therefore, uₓ = u * cosθ, where θ is the angle of projection.
- The vertical component, uᵧ, changes due to the acceleration due to gravity. Therefore, uᵧ = u * sinθ.

2. The time of flight, T, can be calculated using the equation of motion:
- h = uᵧ * t - 0.5 * g * t²

3. We can solve this equation for T by substituting the given values of h, uᵧ, g, and t. This will give us a quadratic equation in T.

4. The horizontal range, R, can be calculated using the equation of motion:
- R = uₓ * T

5. We can solve this equation for R by substituting the given values of uₓ and T.

6. Finally, we can find the angle of projection θ by using the equation:
- θ = tan⁻¹(R/h)

**Calculations:**

1. Finding uₓ and uᵧ:
- uₓ = 50 * cosθ
- uᵧ = 50 * sinθ

2. Finding T:
- h = uᵧ * t - 0.5 * g * t²
- 20 = (50 * sinθ) * 4 - 0.5 * 9.8 * 4²
- 20 = 200 * sinθ - 78.4
- 200 * sinθ = 98.4
- sinθ = 0.492

3. Finding R:
- R = uₓ * T
- R = (50 * cosθ) * T

4. Substituting sinθ = 0.492 and solving for R:
- R = (50 * cos⁻¹(0.492)) * T

5. Substituting T = 4 seconds and solving for R:
- R = (50 * cos⁻¹(0.492)) * 4

6. Finally, finding the angle of projection θ:
- θ = tan⁻¹(R/h)
- θ = tan⁻¹(((50 * cos⁻¹(0.492)) * 4)/20)

**Conclusion:**

By solving the above equation, we can find the angle of projection of the particle.
Community Answer
a particle is projected at some angle with velocity 50 metre per secon...
Take vertical motion of the projectile 20=50sin¢×t-1/2gt^2
so, 50sin¢=20/t+1/2gt
=20/4+1/2×10×4
=25
so sin¢=1/2
and ¢=30 degree
answer is 30 degree
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a particle is projected at some angle with velocity 50 metre per second crosses 20m high wall after 4 seconds from the time of projection. find the angle of projection of the particle
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a particle is projected at some angle with velocity 50 metre per second crosses 20m high wall after 4 seconds from the time of projection. find the angle of projection of the particle for Software Development 2025 is part of Software Development preparation. The Question and answers have been prepared according to the Software Development exam syllabus. Information about a particle is projected at some angle with velocity 50 metre per second crosses 20m high wall after 4 seconds from the time of projection. find the angle of projection of the particle covers all topics & solutions for Software Development 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for a particle is projected at some angle with velocity 50 metre per second crosses 20m high wall after 4 seconds from the time of projection. find the angle of projection of the particle.
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