Answer:
A.) 22.12 m/s
B.) 3.27 m
C.) 0.57
Explanation:
Given that a mass of 0.4 kg is dropped from rest in a medium offering a resistance of 0.2|v|, where is measured in meters per second. Acceleration due to gravity is 9.8 m/s^2.
a. If the mass is dropped from a height of 25 m, its initial velocity U will be equal to zero. when it hits the ground, the final velocity V will be:
V^2 = U^2 + 2gH
Substitute g and the height into the formula
V^2 = 2 × 9.8 × 25
V^2 = 490
V = 22.12 m/s
b. If the mass is to attain a velocity of no more than 8 m/s, the maximum height from which it can be dropped will be calculated by using the same formula.
V^2 = U^2 + 2gH
Where V = 8 m/s and U = 0
8^2 = 0 + 2 × 9.8 × H
64 = 19.6H
H = 64/19.6
H = 3.27 m
c. Suppose that the resistive force is k|v|, where v is measured in m/s, and k is constant. If the mass is dropped from a height of 25 m and must hit the ground with a velocity of no more than 8 m/s, determine the coefficient of resistance that is required.
The energy at 25m heigh will be:
E = mgh
E = 0.4 × 9.8 × 25
E = 98 J
1/2mv^2 = 98
1/2 × 0.4 × ( 8 - v )^2 = 98
0.2( 8 - v )^2 = 98
( 8-v)^2 = 98/0.2
( 8 - v )^2 = 490
8 - v = 22.12
v = 22.14 - 8
v = 14.14
Coefficient of resistance required will be 8/ 14.14 = 0.57
