Answer:
C. 30.6m
Explanation:
To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.
Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

Answer:
3 seconds
Explanation:
Applying,
Applying,
v = u±gt................ Equation 1
Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.
From the question,
Given: v = 0 m/s ( at the maximum height), u = 30 m/s
Constant: g = -10 m/s
Substitute these values into equation 1
0 = 30-10t
10t = 30
t = 30/10
t = 3 seconds
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr
The Answer is:
O 3s
Hope you got it right.