A projectile is fired horizontally with a velocityu . Obtain the expression for speed of theparticle when it hits the ground .

Explanation

When the projectile hits the ground after initially thrown horizontally from the top of tower of height h . The time of flight is given by
`t = sqrt((2h)/(g))`
the horizontal component velocity of the projectile remains the same i.e.,` v_(x) = u`
Hence the vertical component velocity of the projectile at time T is given by
` v_(y) = gT = g sqrt((2h)/(g)) = sqrt(2gh)`
The speed of the particle when it reaches the ground is
`v = sqrt (u^(2) + 2 g h)` .