Question

Let three balls 1, 2, and 3 are allowed to fall under gravity from the same
height. Ball 1 is thrown vertically upward with speed u and it reaches the
ground in time t1, Ball 2 is thrown vertically downward with the
same speed u and it reaches the ground in time t2, Ball 3 is dropped (i.e., u =
O) from the same height and it reaches ground in time t3 .

PROVE that the relationship
between t1, t2, and t3, is given by,
t3 √ t1t2

Answer 1

case 1 : Ball 1 is thrown vertically upward from height h, with speed u m/s. time taken by Ball 1 to reach the ground is t1.

using formula, S = ut + 1/2 at²

here, S = -h , t = t1 and a = -g

so, -h = ut1 -1/2 gt1²

or, -2h = 2ut1 - gt1²

or, gt1² - 2ut1 - 2h = 0

or, t1 = {2u ± √(4u² + 8gh)}/2g

but, t1 ≠ {u - √(u² + 2gh)}/g because it is negative value.

so, t1 = {u + √(u² + 2gh)}/g .....(1)

case 2 : Ball 2 is thrown vertically downward from height h, with speed u m/s

time taken to reach the ground by Ball 2 is t2 .

so, using S = -h, u = -u and a = -g

then, -h =-ut2 - 1/2gt2²

or, -2h = -2ut2 - gt2²

or, gt2² + 2ut2 - 2h = 0

or, t2 = {-2u ± √(4u² + 8gh)}/2g

or, t2 = {-u ± √(u² + 2gh)}/g

but t2 ≠ {-u - √(u² + 2gh)}/g

so, t2 = {-u + √(u² + 2gh)}/g......(2)

case 3 : Ball 3 is dropped from height h,

Time taken to reach the ground by Ball 3 is t3

so, -h = 0 + 1/2 × -g × t3²

or, t3 = √{2h/g}......(3)

multiplying equation (1) and (2),

t1t2 = {u² + 2gh - u²}/g² = 2h/g

or, √t1t2 = √{2h/g} = t3 [ from equation (3) ]

hence, proved