Question

you throw one ball with a initial velocity of "u" vertically up and wait "t" seconds and throw a second ball with the same velocity as the first ball. prove that the both balls will collide after (1/2+u/g) seconds later (gravity is g)

Answer 1

The two balls will collide in
`((t/2) + (u / g))` seconds.

Step-by-step explanation:

Assuming that air resistance is negligible, the vertical acceleration of both balls would be
`a = (-g)`, where
`g` is the gravitational field strength.

Let
`T` denote the number of seconds after the first ball was launched. Apply the SUVAT equation to find an expression for the height
`h_(1)` of the first ball as a function of
`T\!`:

`\begin{aligned} h_(1)(T) &= (1)/(2)\, a\, T^(2) + u\, T \\ &= (1)/(2)\, (-g)\, T^(2) + u\, T\end{aligned}`.

The SUVAT equation above gives the height of the ball at a given time after the ball was launched. For the first ball, that time is equal to
`T`. However, because the second ball was launched
`t` seconds after the first, the second ball would be in the air for only
`(T - t)` seconds at time
`T\!` after the first ball was launched.

Hence, the height of the second ball at time
`T` after the first ball was launched would be:

`\begin{aligned} h_(2)(T) &= (1)/(2)\, a\, (T - t)^(2) + u\, (T - t) \\ &= (1)/(2)\, (-g)\, (T - t)^(2) + u\, (T - t) \\ &= (1)/(2)\, (-g)\, T^(2) + g\, t\, T + (1)/(2)\, (-g)\, t^(2) + u\, T - u\, t \\ &= \left((1)/(2)\, (-g)\, T^(2) + u\, T\right) + g\, t\, T + (1)/(2)\, (-g)\, t^(2) - u\, t \end{aligned}`.

To find the time when the heights of the two balls are the same, set
`h_(1)(T) = h_(2)(T)` and solve for
`T`:

`h_(1)(T) = h_(2)(T)`.

`\displaystyle(1)/(2)\, (-g)\, T^(2) + u\, T = \left((1)/(2)\, (-g)\, T^(2) + u\, T\right) + g\, t\, T + (1)/(2)\, (-g)\, t^(2) - u\, t \end{aligned}`.

`\displaystyle g\, t\, T + (1)/(2)\, (-g)\, t^(2) - u\, t = 0`.

`\begin{aligned}g\, t\, T = (1)/(2)\, g\, t^(2) + u\, t \end{aligned}`.

`\begin{aligned}T &= (1)/(g\, t)\, \left((1)/(2)\, g\, t^(2) + u\, t\right) \\ &= (g\, t^(2))/(2\, g\, t) + (u\, t)/(g\, t) \\ &= (t)/(2) + (u)/(g)\end{aligned}`.

In other words, the two ball will collide (heights of the two ball are the same) at
`((t/2) + (u / g))` seconds after the first ball was launched.