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.
Answer:
https://graphicdesign.stackexchange.com/questions/8547/how-to-find-specific-vector-images-like-this-one
This may be wrong but i would put 48, 78 and 118
Answer: it should be 565.5
Step-by-step explanation:
The correct model of the height of rocket above water is;
h(t) = -16t² + 96t + 112
Answer:
time to reach max height = 3 seconds
h_max = 256 ft
Time to hit the water = 7 seconds
Step-by-step explanation:
We are given height of water above rocket;
h(t) = -16t² + 96t + 112
From labeling quadratic equations, we know that from the equation given, we have;
a = -16 and b = 96 and c = 112
To find the time to reach maximum height, we will use the vertex formula which is; -b/2a
t_max = -96/(2 × -16)
t_max = 3 seconds
Thus, maximum height will be at t = 3 secs
Thus;
h_max = h(3) = -16(3)² + 96(3) + 112
h_max = -144 + 288 + 112
h_max = 256 ft
Time for it to hit the water means that height is zero.
Thus;
-16t² + 96t + 112 = 0
From online quadratic formula, we have;
t = 7 seconds
