Physics Energy - how fast ball is moving from certain height

[nasu]

This is v^2. You did not read the previous post?
How do you find v if v^2 is 2.352?

 

[astru025]

Yes so I take the square root of 2.532 to get what v is? Which would be 1.5

 

[nasu]

Yes, take the square root.

 

[astru025]

Ugh I entered this and it said it was incorrect. Is 1.5 m/s not the right answer?!

 

[nasu]

It may be a rounding error.
Or it may be a little more complicated. Depends on the level of your class.

Have you learned about moment of inertia and rotational kinetic energy?
They say that the ball is rolling so you may be supposed to consider the rotation of the ball too.
In this case the kinetic energy will have two terms,
KE=1/2mv^2 + 1/2 I ω^2
where I is the moment of inertia of the ball and ω is the angular speed. For rolling without slipping, ω is v/R where R is the radius of the ball.
Have you seen any of these?

 

[NihalSh]

It may be a rounding error.
Or it may be a little more complicated. Depends on the level of your class.

Have you learned about moment of inertia and rotational kinetic energy?
They say that the ball is rolling so you may be supposed to consider the rotation of the ball too.
In this case the kinetic energy will have two terms,
KE=1/2mv^2 + 1/2 I ω^2
where I is the moment of inertia of the ball and ω is the angular speed. For rolling without slipping, ω is v/R where R is the radius of the ball.
Have you seen any of these?

The question specifically mentions that the ball rolls. That means:

change in gravitational potential energy = (translation kinetic energy)+(rotational kinetic energy)

assuming ball is a solid sphere, you should know the moment of inertia of sphere. and as nasu already mentioned for pure rolling $ω= \frac{v}{r}$
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iω^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\frac{2mr^2}{5}ω^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\frac{2mr^2}{5}(\frac{v}{r})^2$

the final equation would come out to be $mgh = \frac{7}{10}mv^2$