Hello, Ashley!

I *think* I have this solved . . .

A girl stands at a waters edge and throws a stone from a height of 0.9 m.

To get the stone to bounce, it must hit the water at an angle of 15°.

Find the minimum speed which she can throw the stone.

Well, i calculated the time for the stone to fall: 0.429 seconds (correct in the answers).

The vertical component of the stone hitting the water, the speed is 4.2 m/s. Assuming she throws the stone *horizontally*, your calculations are correct.

As you found, the vertical component of the speed is: $\displaystyle 4.2$ m/sec.

Let the original (horizontal) speed be $\displaystyle v$. .This is constant throughout the flight.

When the stone strikes the water, we have this diagram: Code:

v
* - - - - - - - - - - - *
* 15° |
* |
* | 4.2
* |
* |
*

In the right triangle: .$\displaystyle \tan15^o \:=\:\frac{4.2}{v} \quad\Rightarrow\quad v \:=\:\frac{4,2}{\tan15^o} \:=\:15.67461339$

Therefore: .$\displaystyle v \:\approx\:15.67$ m/sec.