Question

Salmon often jump waterfalls to reach their breeding grounds. Starting 3.15 m from a waterfall 0.45 m in height, at what minimum speed must a salmon jumping at an angle of 33.2 degree leave the water to continue upstream? the acceleration due to gravity is 9.81 m/s2. Answer in units of m/s.

Answer 1

Let the salmon jump with speed v at an angle of 33.3 degree

So here we can find out the components of his velocity in x and y direction

$v_x = v cos33.2$

$v_y = v sin33.2$

now the horizontal displacement of the salmon is 3.15 m so he cover this horizontal distance with constant speed as there is no acceleration in x direction

here we can say

$v_x * t = 3.15$

$v cos33.2 * t = 3.15$

$v*t = 3.76$

now in Y direction it is an accelerated motion as it is accelerated due to gravity

$y = v_y * t + \frac{1}{2}at^2$

$-0.45 = v sin33.2* t - \frac{1}{2}*9.8* t^2$

now we will plug in v*t = 3.76

$-0.45 = 3.76 * sin33.2 - 4.9 * t^2$

$4.9 t^2 = 2.51$

$t^2 = 0.512$

$t = 0.72 s$

now speed v is given by equation above

$v*t = 3.76$

$v*0.72 = 3.76$

$v = 5.25 m/s$

so here he must have to jump with minimum speed of 5.25 m/s