A diver makes 2.5 revolutions on the way from a 10 -m-high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is

Question

A diver makes 2.5 revolutions on the way from a 10 -m-high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is (A) (3 π/√2) rad / s (B) (5 π/√2) rad / s (C) (5 π/√3) rad / s (D) (π/√2) rad / s

Tardigrade Admin · Accepted Answer

The free-fall time: Δ y=v0 y t+(1/2) g t2 ⇒ t=√(2(10 m )/10 m / s 2)=√2 s Thus, the magnitude of the average angular velocity is ω avg =((2.5 rev )(2 π rad / rev )/√2 s )=(5 π/√2) rad / s

Q. A diver makes $2.5$ revolutions on the way from a 10 -m-high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is

$\frac{3 π}{2} r a d / s$

$\frac{5 π}{2} r a d / s$

$\frac{5 π}{3} r a d / s$

$\frac{π}{2} r a d / s$

The free-fall time:
$Δ y = v_{0 y} t + \frac{1}{2} g t^{2}$
$\Rightarrow t = \frac{2 ( 10 m )}{10 m / s ^{2}} = 2 s$
Thus, the magnitude of the average angular velocity is
$ω_{a vg} = \frac{( 2.5 re v ) ( 2 π r a d / re v )}{2 s} = \frac{5 π}{2} r a d / s$

Q. A diver makes 2.5 revolutions on the way from a 10 -m-high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is

2​3π​rad/s

2​5π​rad/s

3​5π​rad/s

2​π​rad/s