Question

An object falls from a height h from rest. If it travels a fraction of the total height of 0.5833 in the last 1.00 s, find the time of its fall.

Find the height of its fall

Answer 1

<< the height of its fall >>

Let

h = total height from which the ball was dropped

The first working formula is

0.5833h = VT + (1/2)(gT^2)

where

V = velocity of the ball at a height of (0.5833h) from the ground
T = time = 1 second (given)
g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

(0.5833h) = (V)(1) + (1/2)(9.8)(1)

0.5833h = V + 4.9

Solving for "V",

V = 0.5833h - 4.9 --- call this Equation 1

The next working formula to used, relating V and H, is

V^2 - Vo^2 = 2(g)(0.5833h)

where

Vo = initial velocity = 0 (since ball was dropped)

and all the other terms have been previously defined.

Solving for V^2,

V^2 = (2)(9.8)(0.5833h) = 11.43h

Substituting Equation 1 in the above,

(0.5833h - 4.9)^2 = 11.43h

Simplifying the above,

0.34h^2 - 2.85h + 24.01 = 9.8h

0.34h^2 - 12.65h + 24.01 = 0

Using the quadratic formula,

h = 2.006 and h = 35.19

Mathematically, there are 2 possible values of "h" that will satisfy the above quadratic equation.


<< the total time of its fall >>

Working formula is

h = VoT + 1/2(g)T^2

where

T = total time of fall

and all the other terms have been previously defined.

Solving for T,

T = sqrt (2h/g)

For h = 35.19

T = sqrt (2*35.19/9.8)

T = 2.67 seconds


For h = 2.006

T = sqrt (2*2.006/9.8)

T = 0.639 sec.

Since T = 0.639 sec contradicts one of the given conditions of the problem (travels 0.5833h in the last 1 sec), then the root h = 2.006 is not a valid answer.

For this particular problem,

h = 35.19 meters

and

T = 2.67 sec.