Question

A body is projected vertically up with velocity ‘u’ and ‘t’ seconds later, another body is projected vertically upwards from the same point and with the same velocity. They will meet at a height​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• They meet at the height

$\:\:\:\:\:\:\:\:\:\:\red{\boxed{ h = \left(\frac{u^2}{2g} - \frac{gt^2}{8}\right)}}$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Let us suppose that the two body meets $\Delta t$ sec after the second body is thrown up, at height 'h'.

For the first body

$u = u, a = - g$

$S = h, Time = (t+\Delta t)$

Using equation of motion,

$S = ut + \frac{1}{2}at^2$

$h = u(t + \Delta t) - \frac{1}{2}g(t +\Delta t)^2... (1)$

For the second body

$u = u, a = - g$

$S = h, Time = \Delta t$

Using equation of motion,

$S = ut + \frac{1}{2}at^2$

$h = u(\Delta t) - \frac{1}{2}g(\Delta t)^2... (2)$

Substracting (1) and (2), we get

$0 = ut - \frac{1}{2}g[(t +\Delta t)^2-(\Delta t)^2]$

$ut = \frac{1}{2}g[t^2 +2t\Delta t]$

$\frac{2ut}{g} = t^2 +2t\Delta t$

$\frac{2ut}{g} - t^2 = 2t\Delta t$

$\frac{2u}{g} - t = 2\Delta t$

$\to \boxed{\Delta t = \left(\frac{u}{g} - \frac{t}{2}\right)}$

Substituting this value in (2), we get

$h = u\left(\frac{u}{g} - \frac{t}{2}\right) - \frac{1}{2}g\left(\frac{u}{g} - \frac{t}{2}\right)^2$

$h = \left(\frac{u^2}{g} - \frac{ut}{2}\right) - \frac{1}{2}g\left(\frac{u^2}{g^2} -2\frac{u}{g}\frac{t}{2}+ \frac{t^2}{4}\right)$

$h = \left(\frac{u^2}{g} - \frac{ut}{2}\right) - \left(\frac{u^2}{2g} -\frac{ut}{2}+ \frac{gt^2}{8}\right)$

$\to \boxed{ h = \left(\frac{u^2}{2g} - \frac{gt^2}{8}\right)}$