Dropping a ball: height and time??

Question

A ball, dropped from rest, covers five-sevenths of the distance to the ground in the last two seconds of its fall.

(a) From what height was the ball dropped?

(b) What was the total time of fall?

Answer 1

a)h= 0.5 gt^2

since it is 5/7 of total h in 2 s

and assuming metric system

h'=  0.5 x 9.81 x 2^2

h'= 19.6 m

then

h=( 7/5)h' = 27.5 m

b) Since t=sqrt( 2 h /g)

t= 2.37s

Answer 2

<< From what height was the ball dropped? >>

Let

H = total height from which the ball was dropped

The first working formula is

(5/7)H = VT + (1/2)(gT^2)

where

V =  velocity of the ball at a height of (5H/7) from the ground

T = time = 2 seconds (given)

g =  acceleration due to gravity

Substituting appropriate values,

(5H/7) = (V)(2) + (1/2)(9.8)(4)

5H/7 = 2V + 19.6

Solving for "V",

V = (5H/14) - 9.8  --- call this Equation 1

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

V^2 - Vo^2 = 2(g)(2H/7)

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)(2H/7) = 5.6H

Substituting Equation 1 in the above,

[(5H/14) - 9.8]^2 = 5.6H

Simplifying the above,

25H^2/196 - (9.8)(5)H/7 + 96.04 = 5.6H

25H^2/196 - 12.6H + 96.04 = 0

Using the quadratic formula,

H = 8.32 and H = 90.46

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



<< What was the total time of 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 = 90.46,

T = sqrt (2*90.46/9.8)

T = 4.30 seconds

For H = 8.32

T = sqrt (2*8.32/9.8)

T = 1.30 sec.

Since T = 1.30 sec will not satisfy the conditions of the problem, then the root H = 8.32 is not a valid answer.

For this particular problem,

H = 90.46 meters

and

T = 4.30 sec.

Hope this helps.