Answer:
The period when its more than halfway
= 2 < t ≤ 2.83 seconds ( to nearest hundredth).
Step-by-step explanation:
When is is halfway down the ramp d = 3920/2 = 1960 cm, so:
1960 = 490 t^2
t^2 = 1960/490 = 4
t = 2 seconds.
When it reaches the bottom d = 3920 :
t^2 = 3920/490 = 8
t = 2.83 seconds.