Physics, 11.10.2019 04:10 kayleg907436
Astone is thrown with an initial velocity of 36 ft/s from the edge of a bridge that is 46 ft above the ground. the height of this stone above the ground t seconds after it is thrown is f(t)= -16t^2 + 36t + 46. if a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t)= -16t^2 + v0t, where v0 is the initial velocity of the second stone. determine the value of v0 such that the two stones reach the same high point.
Answers: 3
is the initial velocity of the stone, g is the gravity and 
is the initial position of the stone.
, so g is equal to 
of the first stone is:
as:
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Astone is thrown with an initial velocity of 36 ft/s from the edge of a bridge that is 46 ft above t...
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