Posted by **just wondering** on Monday, February 21, 2011 at 3:34pm.

A flower pot is dropped from the balconey of an apartment 28.5 metres above the ground. At a time of 1 second after the pot is dropped a ball is thrown vertically downwards from the balconey one story below 26 metres above the ground. The initial velocity of the ball is 12 metres per second down. Does the ball pass the flowerpot before striking the ground? If so how far above the ground are the two objects when the ball passes the flower pot?

Also please explain step to step how to come to the conclusion of this question

- physics gr 12 -
**Damon**, Monday, February 21, 2011 at 4:33pm
ball in air for time t so pot in air for (t+1)

flower pot

h = 28.5 - 4.9 (t+1)^2

ball

h = 26 - 12 t - 4.9 t^2

when are they at the same h ?

26-12 t -4.9 t^2 = 28.5 - 4.9 (t^2+2t+1)

-2.5 -12 t = -9.8 t -4.9

2.4 = 2.2 t

t = 1.1 seconds for ball in air

and

2.1 seconds for pot in air

now h

h ball = 26 - 12 (1.1) - 4.9 (1.1)^2

= 26 - 13.2 - 5.9

= 6.9 meters from ground so they do pass before hitting ground

check height of pot

h = 28.5 - 4.9 (2.1)^2

=28.5 - 21.6

= 6.9 meters again !! Good

- physics gr 12 -
**just wondering**, Wednesday, February 23, 2011 at 6:33pm
im confused about how u used the difference in time to solve it?

- physics gr 12 -
**Daniel**, Wednesday, October 1, 2014 at 9:51pm
The reply by Damon is wrong.

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