Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 `ms^(- )`and 30 `ms^(-1).` Verify that the graph shown in figure. Correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 `ms^(-2).` Give the equations for the linear and curved parts of the plot.

Explanation

`x _(0) = 200 m, u = 15 ms ^(-1) , a =- 10 ms ^(-2) and x = x _(1)`
`therefore` From equation of motion.
`x =x _(0) + ut + 1/2 at ^(2)`
`therefore x _(1) = 200 + 15 t + 1/2 xx (-10) t ^(2)`
`therefore x_(1) = 200 + 15t - 5t ^(2) " "...(1)`
When first stone strikes the ground, `x _(1) =0`
`therefore 0=40 + 3t -t ^(2)`
`therefore t ^(2) - 3t - 4=0`
`therefore (t-8) (t +5) =0`
`therefore t = 8 s or t =-5s`
But `t =-5` not possible.
`therefore t=8s`
For second stone,
`x_(0) = 200 m, u = 300ms ^(-1), a =- 10 ms ^(-2) and x=x_(2)`
`therefore x_(2) = 200 + 30 t + 1/2 xx (-10)t ^(2)`
` therefore x_(2) = 200 + 30 t - 5t ^(2) " "..(2)`
When second stone strikes the ground, `x _(2) =0`
`therefore 0=200 + 30t - 5t ^(2)`
`therefore 5t^(2) - 30 t - 200=0`
`therefore t ^(2) - 6t - 40=0`
`therefore (t-10) (t+4) =0`
`therefore t = 10 s and t =-4s`
But `t =-4s` not possible.
`therefore t = 10s`
By substracting equation (1) From (2),
`x _(2) - x _(1) = 200 + 30 t - 5t ^(2) - 200 - 15 t + 5t ^(2)`
`therefore x _(2) -x_(1) = 15 t` (Linear part) ...(3)
Relative velocity of stone,
`u _(21) = u _(2) - u _(1)`
`= 30 -15`
`therefore u _(21) = 15 ms ^(-1)`
Equation (3) is equation of a line. Hence graph of `(x_(2)-x_(1))` + t will be linear upto OA.

After 8s, first stone will be on ground and only second stone will be in motion.
`therefore` Maximum distance of both stones after 8 s,
`(x_(2) - x _(1)) = 15 t`
`= 15 xx 8 = 120 m`
And after 8 s, second store will strike the ground by performing hyperbolic motion,
`therefore AB ` path will be hyperbolic,
`x_(2) - x _(1) = x_(0) + u _(21) t + 1/2 at ^(2)`
`= 200 + 15t + 1/2 xx (-10) t ^(2) = 200 + 15t - 5t ^(2)`
Hence, the above equation is for AB part of graph (curved part).