hi me with this question. .
a man throws a stone upwards with a 'u' velocity and after spending ' t' seconds  watching the motion of the stone he once again projects another stone upwards hoping that that these 2 stones should  meet eachother.  if his assumption is correct and these two stones collide with eachother,  show that the 2 stones should meet each other 't/2 + u/g 'seconds later from the time of the first projection.

Explanation:

For projectile motion, use constant acceleration equation:

Δx = v₀ t + ½ at²

where Δx is the displacement,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Both objects are projected upward with velocity u.  The second object is thrown after a time t₀.

For the first object:

Δx = u t + ½ (-g) t²

Δx = u t − ½g t²

For the second object:

Δx = u (t−t₀) + ½ (-g) (t−t₀)²

Δx = u (t−t₀) − ½g (t−t₀)²

Assuming the objects meet, the displacements will be equal:

u t − ½g t² = u (t−t₀) − ½g (t−t₀)²

u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)

u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²

0 = -u t₀ + g tt₀ − ½g t₀²

0 = -u + g t − ½g t₀

g t = u + ½g t₀

t = u/g + t₀/2

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