What is the minimum distance she can be from the waterfall?

[kmikias]

Hi.
I have question for you guys...i tryed but i couldn't
here is the question.

1.A salmon is swimming upstream to return to its birthplace in order to mate.along the way,she encounters a small waterfall 0.8m tall.she jumps out of the water with a velocity of 5m/s.

A) what is the minimum distance she can be from the waterfall?
B)At what angle must she leave the water?

Here is what i did
here velocity is 5m/s and vertical distnce will be 0.8m and gravity will be 9.8

I use verrtical and horizontal formula which is
y=vt+1/2gt^2
and X=vt

here is where i couldn't solve
how can i find velocity of y and x without knowing the angle .
i know velocity of y=sin(angle) *5m/s
X=cos(angle)5m/s
AND i don't even have the time

thanks.

 

[Almanzo]

To jump to a certain height, she must start out with a certain kinetic energy, equal to the potential energy she will have on reaching that height. Her mass is unimportant, because if (1/2)mv² = mgh, then (1/2)v² = gh

Happily, the kinetic energy for vertical and horizontal motion add, by Pythagoras.
v_x²+v_y²=v².

So you calculate the total energy, and you calculate the fraction of that energy needed to reach the desired height, and the square root of that fraction will be the sine of the angle she must jump. The square root of the fraction not needed will be the cosine. The square root of their quotient will be the tangent.

5 m/s squared and halved gives 12.5 m²/s².
0.8 m height times 10 m/s² (the acceleration of gravity) gives 8 m²/s².

So the fraction needed is 16/25, the fraction not needed is therefore 9/25.
The sine is 4/5, the cosine is 3/5, the tangent is 4/3. From this the angle can be found in a table.

If a gravity of 9.81 m/s² is to be used, things become more complex, but essentially similar.

If she is just clearing the fall her mean vertical speed will be half her initial vertical speed. Her vertical speed has linearly decreased to zero from start to end. So she will reach the height in twice the time she would need to reach it if no gravity existed. From this we see that the horizontal distance covered in that time will be twice the cotangent multiplied by the height. The cotangent is 3/4, twice that is 3/2, and the maximum distance is 1.2 meter.