The discussion focuses on calculating the distance of two free-falling objects dropped from a height of 60 meters, with initial velocities of 20 m/s and 15 m/s, respectively. The kinematic equation used is D = iV * T + (1/2) * a * t^2, where 'D' is the distance, 'iV' is the initial velocity, 'a' is the acceleration due to gravity (approximated as 10 m/s²), and 'T' is the time. The first object reaches the ground in 2 seconds, while the second object, using the same formula with its initial velocity, is determined to be at a distance of 10 meters when the first object hits the ground.
PREREQUISITESStudents studying physics, educators teaching kinematics, and anyone interested in understanding the principles of free fall and motion under gravity.
Good. So how can you apply this to solve for the time it takes for the 20 m/s object to reach the ground?Purpplehaze said:D= iV * T + a*t^2/2 I think that's the formula to be used.
Purpplehaze said:D= iV * T + a*t^2/2 I think that's the formula to be used.
Purpplehaze said:Would this work? 2(D-iV)/a = t
No.Purpplehaze said:Would this work? 2(D-iV)/a = t
Purpplehaze said:Then how can I isolate t?
Purpplehaze said:60=20t + 4.9t^2
Good. Now you can put it into standard form and solve it using the quadratic formula (or using whatever method you like).Purpplehaze said:60=20t + 4.9t^2
Good!Purpplehaze said:I got 2 as T for the first one.