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## Homework Statement

A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m at 30 degrees above the horizontal.

What is the maximum height the stone reaches above the water? (ans:42.87m)

## Homework Equations

**Vfx=VicosΘ****Δx=vicosΘ(t)****Vfy=(VicosΘ)+ay(t)****Δy=(VisinΘ)(t)+1/2(ay)(t)****(Vfy)^2=(VisinΘ)^2+2(ay)(Δy)****Δy=1/2(Vfy+(VisinΘ))(t)**

## The Attempt at a Solution

ax:13

ay:7.5

Δy:40m

Δx:48m[/B]

I found the x and y-components.

__Ax:15cos30=13__and

__Ay:15sin30=7.5.__

Then I tried to find the time using the second equation and got

__3.69s.__

I've been trying to solve for vfy.... and I've done everything. I'm not getting that answer. I don't know why I can't understand how to do this. it's frustrating... I even tried to draw it on an axis.