fisica1-youn-e-freedman-exercicios-resolvidos

2.98: a) Let the height be h and denote the 1.30-s interval as ∆ t; the simultaneous
2 2 1
2
equations h =
1 gt , h = g(
t − ∆t
) can be solved for t. Eliminating h and taking the
2
3
2
t 3
∆t
square root, = , andt , and substitution into
=
t − ∆t
2
1−
2 3
2
h = 1
gt
2
gives h = 246 m.
This method avoids use of the quadratic formula; the quadratic formula is a
generalization of the method of “completing the square”, and in the above form,
2 1
2
h = g(
t − ∆ t ) , the square is already completed.
3
2
b) The above method assumed that t >0 when the square root was taken. The
negative root (with ∆t = 0) gives an answer of 2.51 m, clearly not a “cliff”. This would
correspond to an object that was initially near the bottom of this “cliff” being thrown
upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically
possible, the conditions of the problem preclude this answer.