Question

A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown( one upwards and one downwards ) with the same velocity u and they reach the earth surface after t1 and t3 seconds respectively , then
1) t= t1 - t2
2) t = t1+ t2 ÷2
3) t=√t1t2

Answer 1

case 1 : A stone is dropped from a building of height h and time taken by stone to reach the ground is t.

then, h = 1/2 gt² ......(1)

case 2 : same stone is thrown vertically downward with speed u m/s. time taken by stone to reach the ground is t1.

so, -h = -ut1 + 1/2 (-g)t1²

h = ut1 + 1/2 gt1²

gt1² + 2ut1 -2h = 0 .......(2)

t1 = {-u + √(u² +2gh)}/g [ t1 ≠ {-u-√(u²+2gh)}/g]

case 3 : same stone is thrown vertically upwards with speed u m/s and time taken by stone to reach the ground is t2.

so, -h = ut2 + 1/2 (-g)t2²

or, h = -ut2 + 1/2 gt2²

or, gt2² - 2ut2 - 2h = 0.......(3)

t2 = {u + √(u² + 2gh)}/g [ t2 ≠ {u - √(u² + 2gh)}/g]

now, t1.t2 =[ {-u + √(u² +2gh)}/g ]. [ {u + √(u² +2gh)}/g ]

= (u² + 2gh - u²)/g²

= 2h/g

from equation (1),

t1.t2 = 2h/g = t²

or, t = √(t1.t2) This is required answer.

Answer 2

Answer:

3)

Solution:

Hope this will help you