Posted by irna on Thursday, April 21, 2011 at 9:53am.
a stone is from the top of a tower 300m high and at the same time another stone is projected vertically upward with initial velocity 75m/s. calculate when and where the two stone meet

physics  Henry, Monday, April 25, 2011 at 11:48am
d1 + d2 = 300m,
0.5gt^2 + (75t + 0.5gt^2) = 300m,
Since g in the 1st term is positive and
g in the 3rd is negative, they cancel
each other and the Eq becomes:
75t = 300,
t = 4 seconds.
d1 = 0.5*9.8*4^2 = 78.4m. below top
of tower = 300  78.4 = 222m above
ground.
d2 = 75*4  0.5*9.8*4^2,
d2 = 300  78.4 = 222m above ground.
Therefore, the 2 stones meet 4s after
release at a distance of 222m above
ground.

physics  Anonymous, Friday, June 10, 2016 at 4:47pm
U suck in explanation
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