Answer:
Explanation:
Model of the frog jump is
h = -6t² + 6t
When will the frog be on the ground? The frog will be on the ground at h=0
Therefore,
0 = -6t² + 6t
Divide through by 6
0 = -t² + t
0 = t(-t + 1)
So, t = 0 OR. 0 = -t + 1
So, t = 0. OR. t = 1
So, we should note that, at t = 0, was when he wanted to lift of from the ground... So, t = 0 will be discarded
So, at t = 1 is when the frog is back to the ground.
It is like it starts from the ground at the 0 seconds and get backs to the ground at 1 seconds
I hope this helps!!!