Throwing marbles

Classical Mechanics Level 1

Two identical marbles A A and B B are thrown vertically from the ground with speeds of 2 m/s 2 \text{ m/s} and 1 m/s , 1 \text{ m/s}, respectively, as shown above. Let a a be the distance traveled by A A until it returns back to the ground. Let b b be the distance traveled by B B until it returns back to the ground. What is a b ? \displaystyle{\frac{a}{b}?}

2 4 6 3

View wiki
Srinivasa Satti by Srinivasa Satti

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mitesh Warang
Jun 11, 2014

v 2 v^{2} = u 2 u^{2} -2gs

velocity at top is zero for both marble

hence, 2gs= u 2 u^{2}

For first height is a= 1/5

For second height is b=1/20

Therefore a/b=4

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I would prefer height of projectile where /theta =90°

Akshay Iyer - 6 years, 12 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...