#### Answer

To have the same speed as the sphere, the hoop should be released from a height of 43 cm

#### Work Step by Step

We can use the formula for the speed of a rotating object to find an expression for the sphere's speed $v_s$ at the bottom. Let $h_s$ be the initial height of the sphere.
$v_s = \sqrt{\frac{2gh_s}{1+c}}$
$v_s = \sqrt{\frac{2gh_s}{1+2/5}}$
$v_s = \sqrt{\frac{10gh_s}{7}}$
We can use the formula for the speed of a rotating object to find an expression for the hoop's speed $v_h$ at the bottom. Let $h_h$ be the initial height of the hoop.
$v_h = \sqrt{\frac{2gh_h}{1+c}}$
$v_h = \sqrt{\frac{2gh_h}{1+1}}$
$v_h = \sqrt{gh_h}$
We then equate the speeds to find the value of $h_h$;
$v_h = v_s$
$\sqrt{gh_h} = \sqrt{\frac{10gh_s}{7}}$
$gh_h = \frac{10gh_s}{7}$
$h_h = \frac{10h_s}{7}$
$h_h = \frac{(10)(0.30~m)}{7}$
$h_h = 0.43~m$
To have the same speed as the sphere, the hoop should be released from a height of 43 cm.