An object falls a distance h from rest. If it travels 0.460h in the last 1.00 s, find (a) the time and (b) the height of its fall.

a) t = 3.771s

b) h = 69.68 m

Explanation:

Given;

total distance covered = h

Using the second equation of motion;

h = ut + 0.5at² ... 1

We know u=0, since the object fall from rest.

and also acceleration a = g (acceleration due to gravity)

The equation 1 becomes;

h = 0.5gt² ... 2

Equation 2 is the overall equation of motion of the object.

Where;

h = height of fall

t = total time of fall

The motion have two phases.

1) distance travelled in the first phase is

h₁ = h - 0.46h = 0.54h

Time taken for the first phase is;

t₁ = t-1

2) distance travelled in the second phase is

h₂ = 0.46h

Time taken for second phase is

t₂ = 1.00s

Writing the equation of motion of the first phase where the object starts from rest, we have;

0.54h = 0.5g (t-1) ² ... 3

Substituting equation 2 into 3, we have;

0.54 (0.5gt²) = 0.5g (t-1) ²

divide both sides by 0.5g

0.54t² = (t-1) ²

(t-1) ² - 0.54t² = 0

Simplifying the equation above, we have;

t² - 2t + 1 - 0.54t² = 0

Then we have a quadratic equation;

0.46t² - 2t + 1 = 0

solving the quadratic equation, we have

t = 3.771 or t = 0.5764

Since t cannot be less than 1,

Recall that t₁ = t-1, and t₁ cannot be negative. so t>1

t = 3.771s

b) Using equation 2

h = 0.5gt²

h = 0.5*9.8*3.771²

h = 69.68 m

A. Time, t = 4.35s

B. Height of its fall, S = 92.72m

Explanation:

Vo = 0 m/s

Vi = 0.46h m/s

S = h m

a = 9.81 m/s2

To calculate the time taken, we need to get the value of the distance, h.

Using the equations of motion,

Vi^2 = Vo^2 + 2aS

Where Vi = final velocity

Vo = initial velocity

a = acceleration due to gravity

S = height of its fall

(0.46h) ^2 = 0 + 2*9.81*h

0.2116h^2 = 19.62h

h = 19.62/0.1648

= 92.722 m

To calculate the time,

S = Vo*t + (1/2) * a*t^2

92.772 = 0 + (1/2) * 9.81*t^2

t^2 = 185.44/9.81

= 18.904

t = sqrt (18.904)

= 4.348 s