The discussion focuses on calculating the distance an object falls during the fourth second of free fall under Earth's gravitational field, assuming negligible air resistance. The relevant equations used include \( v = u + at \) and \( s = ut + \frac{1}{2}at^2 \). The correct answer to the problem is determined to be 35 meters, derived from the total distance traveled at the end of the fourth second minus the distance traveled at the end of the third second. The calculations confirm that the object falls 80 meters in total by the end of the fourth second, leading to the conclusion that the distance traveled during the fourth second is indeed 35 meters.
PREREQUISITESThis discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of kinematics and free fall motion. It is particularly useful for those preparing for exams or solving related homework problems.
could you give me more hint? I do not get it since that there is no total distance...gneill said:
There is a total distance it has traveled from the point of release at time equals zero to any given instant of time after that. One of your equations will give you that distance for any time t.SUSUSUSUSUSUSUSU said:
gneill said:
SUSUSUSUSUSUSUSU said:
gneill said:That's what you want. I'm not sure where your "s= (v+u)t/2" came from, but your second Relevant Equation gives you the correct form directly given that the initial velocity, u, is zero
