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Two particles A and B are projrcted from the same point on the ground. Particle A is projecyed on a smooth horizontal surface with speed v, whereas B is projrcted with velocity 2v/√3. The two particles collide when the particle B drops on the surface for the first time. Find: a) The angle with which the particle B was projected into air. b) The separation between the two particles when B is at its highest position.


2 years ago

Eshan
2095 Points
							Dear student,Since the particle collide on the horizontal surface after finite time, their horizontal displacement must be same. Hence the horizontal component of their velocities must be same. Therefore$\dpi{80} \dfrac{2v}{\sqrt{3}}cos\theta=v\implies cos\theta=\dfrac{\sqrt{3}}{2}\implies \theta=30^{\circ}$When it is at maximum height, the separation is the height of the particle B which is equal to$\dpi{80} \dfrac{u^2sin^2{\theta}}{2g}=\dfrac{v^2}{6g}$

2 years ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions