Answer:
The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>
Step-by-step explanation:
Given:
Initial speed of debris is, 
The height 'h' of the debris above the ground is given as:

As per question,
. Therefore,

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

Using quadratic formula to solve for 't', we get:

Therefore, the debris will reach a height of 56 ft twice.
When time
during the upward journey, the debris is at height of 56 ft.
Again after reaching maximum height, the debris falls back and at
, the height is 56 ft.