Speed of a salmon jumping 3.56m at 45 degrees

[sundrops]

Salmon swimming up a river to their spawning grounds, leap over all sorts of obtacles. The unofficial salmon-altitude record is a 3.56m jump. Assuming that the salmon took off at a 45 degree angle, what was the speed at which it was emerging from the water?


Xo = 0m
Yo = 0m
Y = 3.56m
a = 9.8m/s^2
X = ?
v = ?

and I thought of using this formula : Y = Yo + VYo(t) + (0.5)(a)(t)^2
but I couldn't because I don't have t.

 

[cepheid]

Hint: it's a 45 degree angle, so you should see from a vector diagram that initially v_x = v_y. Calculate the initial vertical velocity required for it to reach a max height of 3.56 metres, and then use that (v0y) to get v0.

Big Hint...if the fish doesn't go any higher than 3.56 metres, then obviously its final velocity (what you have called v) is zero. So now, you want to find the change in vertical velocity. You know the vertical acceleration. You know the maximum height reached. The two unknowns are therefore v0y and t. If you also use the equation v = v0y +at...then you have two equations and two unknowns. The problem is finished.

 

[sundrops]

here's what i did for this problem:

(Vy)^2 = (Vyo)^2 - 2g(y - yo)
0 = (Vyo)^2 - 2(9.8m/s^2)(3.5m-0m)
Vyo = 8.35m/s

my answer was incorrect however, I was wondering where I went wrong...