Anyone good with physics.....???

Question

John is standing on the roof of a building 23.0 m above the ground. He throws a water balloon straight up with an initial velocity of 9.8 m/s. At the instant the water balloon is released, he also drops an open beer bottle falls off the roof. Ignore air resistance.

(a) When does the water balloon hit the ground?

(b) What is the time difference between when the beer bottle hits the ground and the water balloon hits the ground?

(c) What is the speed at which the water balloon hits the ground?

(d) When does the water balloon reach its maximum height?

(e) How high above the ground is the beer bottle when the water balloon is at its maximum height?

m

Answer 1

im glad im done with hs. i hated phys , but now i think its alright....

try this site.

http://www.fearofphysics.com/Xva/xva.htm...

Answer 2

Why the freak do you need to know this?

Answer 3

He just so happens to be throwing the balloon up at the same "velocity" as gravity (yes I know gravity is an acceleration.)  In one second the balloon will stop rising and start falling at 9.8 m/s^2.  Use the formula t^2 = 2d/a to figure out how long it will take for the water balloon to hit the ground.  Balloon goes up for one second at 9.8 m/s, so it's starting out 23 m + 9.8 m above the ground.

The balloon (I'm estimating) will hit the ground anywhere between 1 to 2 seconds after the beer bottle does.

Use vf^2 = vi^2 + 2ad to calculate velocity when it hits the ground.  vi = 0 because it was hanging in the air for a sec when it stopped accelerating upwards and then started accelerating down.

Use the same ideas for d) and e).  Good luck!