Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time

From equation (I)




Now, for maximum height
Put the value of t in equation (I)


(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero




Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
Answer:
E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}
R is radius = Diameter/2 = 0.210m.
At z = 0.2m,
Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,
E can be calculated at distance 0.2m away from the centre of the disk.
Put z = 0.3m and all other values in the equation,
E can be calculated at distance 0.3m away from the centre of the disk