The speed of an object falling from rest is directly proportional to the square root of the distance the objec

Question

The speed of an object falling from rest is directly proportional to the square root of the distance the object has fallen. When an object has fallen 36 ft., its speed is 48 ft/s. How much farther must it fall before its speed is 80 ft/s?

Answer 1

Therefore, speed = k root(distance), where k is a constant
Hence, speed1 / root(distance1) = k = speed2 / root(distance2)

Let x be the distance to be fallen further
Substitute the values into the equation:
48 / root(36) = 80 / root(x+36)
root(x+36) = 80 * 6 / 48
root(x+36) = 10
x + 36 = 100
x = 64 ft

Answer 2

The speed of an object falling from rest is directly proportional to the square root of the distance the object has fallen. When an object has fallen 36 ft., its speed is 48 ft/s. How much farther must it fall before its speed is 80 ft/s?

s = k * SQRT(d)
48 = k * SQRT(36) = 6k, k = 8

80 = 8* SQRT(d)
10 = SQRT(d)
d = 100

additional distance: 100-36=64

Answer 3

No the acceleration is proportional to the distance.

On Earth distance fallen (S)
S=Ut+0.5at^2

Distance = initial velocity (Ut) + half a * t squared

where a is the acceleration due to gravity
and t equals the time the object falls

The distance is given by :
V squared =U sqared + 2as

where V= velocity
U = initial velocity
a = acceleration due to gravity
s = distance fallen

Answer 4

d = 36 ft
s = k√d
48 = k √36
48 = k x 6
k = 8

s = 8√d
80 = 8 √d
10 = √d
d = 100

Will have to fall another 64 ft