Question

please help physics class 11 and spammers will be reported ​

Answer 1

Answer:

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 g(t)² ......(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 = -u(t1) + 1/2 (-g)(t1)²

h = u(t1) + 1/2 g(t1)²

g(t1)² + 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 = u(t2) + 1/2 (-g)(t2)²

or, h = -u(t2) + 1/2 g(t2)²

or, g(t2)² - 2u(t2) - 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.

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