A particle is thrown with velocity u making an angle θ with the vertic...
To solve this problem, we can use the equations of motion for projectile motion. Let's break down the problem step by step.
1. Define the given variables:
- The height of each pole is h.
- The time taken to cross the first pole is 11 seconds.
- The time taken to cross the second pole is 3 seconds.
2. Find the time of flight:
The time of flight, T, is the total time taken by the projectile to reach the ground. Since the particle crosses the top of the second pole in 3 seconds, we can say that the particle takes another 8 seconds to reach the ground. Therefore, T = 11 + 8 = 19 seconds.
3. Find the horizontal component of velocity:
The horizontal component of velocity, uₓ, remains constant throughout the motion. We can find it using the equation uₓ = u * cosθ, where u is the initial velocity and θ is the angle with the vertical. Since the angle is not given, we cannot determine the exact value of uₓ.
4. Find the vertical component of velocity:
The vertical component of velocity, uᵧ, changes due to the effect of gravity. We can find it using the equation uᵧ = u * sinθ.
5. Find the maximum height:
The maximum height, H, can be found using the equation H = (uᵧ² * sin²θ) / (2 * g), where g is the acceleration due to gravity. Since we don't know the exact angle, we cannot calculate the exact value of H.
6. Determine the possible answer options:
Given the options (a) 9.5m, (b) 19.5m, (c) 39.2m, and (d) 49m, we can eliminate option (d) because it is beyond the maximum possible height of a projectile thrown at an angle. We also eliminate option (a) because it is too small compared to the height of the poles. Therefore, the possible answer is either (b) 19.5m or (c) 39.2m.
7. Final answer:
Without knowing the exact angle of projection, we cannot determine the maximum height of the projectile with certainty. However, based on the given options, the most likely answer is (b) 19.5m, as it is closer to the height of the poles.
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