Question

Consider a particle launched at a horizontal velocity v0 from a height h above the ground. 1. Derive an expression for the time it takes the projectile to strike the ground. Ignore air resistance.

Answer 1

Answer:
`t=\sqrt{(2h)/(g) }`

Explanation: The time taken by the object to reach the ground depends up on the vertical velocity of the object.

The object horizontal velocity component only moves the object in forward direction and does not have any effect vertically

as the object is launched horizontally its initial vertical velocity is zero

`u_y=0`

the object experiences downward fall due to gravitational acceleration only since there is no air resistance

therefore a=g

the distance covered =s=h

now the kinematic equation could be used

`h=u_yt+0.5at^2\\h=0+0.5gt^2\\h=(gt^2)/(2)\\t^2=(2h)/(g) \\t=\sqrt{(2h)/(g) }`