Question

from the top of a tower one stone is dropped another stone is thrown vertically downwards with a velocity of 3 metre per second after how much time will there separation be 10 metres​

Answer 1

Answer:

Hello!

The answer is (-3+209^1/2)/(10)

Explanation:

Let the initial velocity of the first stone be u1 and the second's be u2.

As given-

u1 = 0 m/s

u2 = 3 m/s

Consider their relative initial velocity to be u. It is given by-

u = u2 - u1 = 3 - 0 = 3 m/s downwards.

As we have the relative velocity between the two stones we can use the kinematic equations to obtain the relative distance between them which is given as 10 m.

Using-

s = ut + (1/2)gt^2

10 = 3t + (1/2)(10)t^2

This can be written as-

5t^2 + 3t - 10 = 0

This is a quadratic equation. It can be solved using the formula-

t = [-b +(b^2 - 4ac)^1/2]/2a

t = {-3 +[3^2-4(5)(-10)]^1/2}/2(5)

t = [-3 +(9+200)^1/2]/10

t = [-3 +(209)^1/2]/10

Note-

I have used the symbol ^ to signify the power. t^2 means t squared and t^1/2 means square root of t. I hope you understand.

Be careful with the brackets.