Calculating Horizontal Velocity of Rock Kicked Off Bridge in Arkansas

[MIA6]

Homework Statement

A bridge rises 321m above the Arkansas river. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock's horizontal displacement is 45.0 m. Fine the speed at which the rock was kicked.

Homework Equations

Vi=Vx. Dx=45m. Dy=-321m

The Attempt at a Solution

I know how to find the speed, but my question is on my book, it says that there is no initial vertical velocity. WHy? i think when you kick a rock off the bridge, there is a initial vertical velocity?! so why the vertical initial velocity is 0? hope you can explain it to me.

Thanks.

 

[Astronuc]

The Attempt at a Solution

I know how to find the speed, but my question is on my book, it says that there is no initial vertical velocity. WHy? i think when you kick a rock off the bridge, there is a initial vertical velocity?! so why the vertical initial velocity is 0? hope you can explain it to me.

The rock is kicked horizontally (impulsively) off the bridge so that it has a constant horizontal velocity only as it leaves the bridge. It then starts falling vertically with zero initial velocity, as though it was simply released from rest. It then obviously starts accelerating with gravity.

In the time that it take to fall from the point of release to the ground, it also travels 45 m horizontally.

 

[MIA6]

ok, i get it. so the ball travels 321m vertically in free fall?

 

[Astronuc]

ok, i get it. so the ball travels 321m vertically in free fall?

Yes. Constant acceleration, g, with no initial velocity (and perhaps ignoring wind resistance).

 

[MIA6]

but i have a question, because i know that 321m is a path that vertical straight down, but the path that the ball travels is half of the parabola, so if we meansure the curve and the straight line, they won't be the same.