physics gr 12
posted by just wondering on .
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
ball in air for time t so pot in air for (t+1)
h = 28.5 - 4.9 (t+1)^2
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
2.1 seconds for pot in air
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
im confused about how u used the difference in time to solve it?
The reply by Damon is wrong.