Question

A rock thrown horizontally from a bridge. Show that the height of the bridge hits the water below. The rock travels in a smooth parabolic path in time. Show that is 1/2gt^2

Answer 1

This situation is related to parabolic motion and the main equation is:

`y=y_(o)+V_(oy) t-(gt^(2))/(2)` (1)

Where:

`y=0` is the final height of the rock, asuming the top of the bridge touches the surface of the water

`y_(o)` is the initial height of the rock

`V_(oy)=0` is the vertical component of the initial velocity (it is zero because the rock was thrown horizontally)

`t` is the time the parabolic motion lasts

`g` is the acceleration due gravity

Rewritting (1) with these conditions:

`0=y_(o)+(0) t-(gt^(2))/(2)` (2)

Hence:

`y_(o)=(gt^(2))/(2)`