An object falls from height h from rest. If it travels .50h in the last 1.00s, find...?

Question

(A) the time and (B) the height of its fall. Explain the physically unacceptable solution of the quadratic equation in t that you obtain.

Answer 1

<< the height of its fall  >>

Let

h = total height from which the ball was dropped

The first working formula is

0.5h = VT + (1/2)(gT^2)

where

V = velocity of the ball at a height of (0.5h) from the ground

T = time = 1 second (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

(0.5h) = (V)(1) + (1/2)(9.8)(1)

0.5h = V + 4.9

Solving for "V",

V = 0.5h - 4.9 --- call this Equation 1

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

V^2 - Vo^2 = 2(g)(0.5h)

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)(0.5h) = 9.8h

Substituting Equation 1 in the above,

(0.5h - 4.9)^2 = 9.8h

Simplifying the above,

0.25h^2 - 4.9h +  24.01 = 9.8h

0.25h^2 - 14.7h + 24.01 = 0

Using the quadratic formula,

h = 1.68 and h = 57.12

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 = 57.12,

T = sqrt (2*57.12/9.8)

T = 3.414 seconds

For h = 1.68

T = sqrt (2*1.68/9.8)

T = 0.586 sec.

Since T = 0.586 sec contradicts one of the given conditions of the problem (travels 0.5h in the last 1 sec), then the root h = 1.68 is not a valid answer.

For this particular problem,

h = 57.12 meters

and

T = 3.414 sec.

Hope this helps.