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

Answer:

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

$y=y_{o}+V_{oy} t-\frac{gt^{2}}{2}$   (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:

$0=y_{o}+(0) t-\frac{gt^{2}}{2}$   (2)

Hence:

$y_{o}=\frac{gt^{2}}{2}$