B) h/( ty2 – tx2)
C) h2/( tx - ty)
D) 2h/( ty2 – tx2)
Solution:
To obtain the displacement, we have to integrate the eq ④
With reference to the diagram above, to compute the acceleration, g, between time tx and ty, the time taken t = ( ty - tx), displacement y = h, and initial velocity Vi = Vx
Substituting into eq ⑤
Supporting Explanations
Source: http://www.school-for-champions.com/science/gravity_equations_derivation.htm
Overview of distance and time relationships
The displacement a falling object travels in a given time is found by knowing that velocity is the change in displacement with respect to time:
v = dy/dt
Substituting for v in the equation v = gt + vi and integrating, we get:
y = gt2/2 + vit
Note: Sincethe convention is that down is a positive direction, the downward distance y from the starting point is also positive.
Rearranging y = gt2/2 + vit and solving the quadratic equation for t gives you:
t = [ −vi± √(vi2+ 2gy) ]/g
This equation can create some confusion because of the plus-or-minus sign. If the object is thrown downward, the plus (+) sign is used. If the object is thrown upward, the sign depends on the object's position with respect to the starting point.
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From: TheSundayTimes, July 4, 2010, Page 7
- MARIS STELLA HIGH SCHOOL, Secondary 4 Preliminary Examination 2, PAPER 1 MULTIPLE CHOICE, NO. 3, 18th September 2008
- http://www.school-for-champions.com/science/gravity_equations_derivation.htm
