JEE Advanced level
A JEE Advanced-level example on Kinematics (Motion in 1D and 2D)
Question. Two particles A and B are projected simultaneously from the same point. Particle A is projected horizontally with speed 20 m/s from a height of 45 m above the ground. Particle B is projected from the ground directly below A at an angle of 53° with the horizontal such that it collides with particle A in mid-air. Find the initial speed of particle B and the time after which they collide. (Take g = 10 m/s², sin 53° = 4/5, cos 53° = 3/5)
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- Set up coordinates with A at origin (height 45 m). For A: x_A = 20t, y_A = -5t². For B starting from ground: x_B = (3u/5)t, y_B = -45 + (4u/5)t - 5t²
- For collision, equate x-coordinates: 20t = (3u/5)t. Since t ≠ 0, we get u = 100/3 m/s
- Equate y-coordinates: -5t² = -45 + (4u/5)t - 5t². This simplifies to 45 = (4u/5)t
- Substitute u = 100/3: 45 = (4/5)(100/3)t = (80/3)t, giving t = 135/80 = 27/16 s
Answer: Initial speed of B = 100/3 m/s (≈ 33.3 m/s), Time of collision = 27/16 s (≈ 1.69 s)
This example was solved twice independently and both solutions agree.