Question

stone projected vertically up from the ground reaches a height y in its path at t1 seconds and for the t2seconds reaches the ground the height y is equals to​

Answer 1

Answer:

The height y is equal to (g/2)(t1 * t2).

Explanation:

As the stone is projected vertically up form the ground, let the initial velocity of the stone be “u”. At “t1” secs the stone reaches a height of “y”. Also, the stone hits the ground after “t2” secs.

Let us consider the displacement of the stone be the height “y” reached by the stone.

So, the total time taken for the complete journey i.e. let the time of flight be  

T = (t1 + t2) sec ….. (i)

We know that,  

T = 2u / g ….. (ii)

Form (i) & (ii), we get

2 u / g = (t1 + t2)

Or, u = g(t1+t2) / 2  

During the acceleration due to gravity is acting downwards on the stone

∴ a = -g.

The height of the stone is given by

S = ut + 1/2gt²

Substituting s = y, a = -g, u = g(t1+t2)/2 and t = t1

Or, y = [(g/2)(t1+t2)t1] + [½ (-g)(t1²)]

Or, y = (g/2) [t1² + (t1 * t2) – t1²]

Or, y = (g/2)(t1 * t2)