A projectile is projected from horizontal ground with velocity 18 km / hr at an angle of 6 0 ∘ from horizontal. Find angular speed (in rad/s) as observed from the point of projection at the time of landing. [ g = 10 m/s 2 ]

ω = r v ⊥ = R u s i n θ = g 2 ( u c o s θ ) ( u s i n θ ) u s i n θ = 2 u c o s θ g = 2 × 5 × 2 1 10 = 2 r a d / s

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PhysicsKinematicsMiscellaneousEasy2 minQB

PhysicsEasynumerical

A projectile is projected from horizontal ground with velocity $18 km / hr$ at an angle of $6 0^{\circ}$ from horizontal. Find angular speed (in rad/s) as observed from the point of projection at the time of landing.

[ g = 10 m/s²]

Answer:

2.00

Solution:

$ω = \frac{v _{⊥}}{r} = \frac{u s i n θ}{R} = \frac{u s i n θ}{\frac{2 ( u c o s θ ) ( u s i n θ )}{g}} = \frac{g}{2 u c o s θ} = \frac{10}{2 \times 5 \times \frac{1}{2}} = 2 r a d / s$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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