Suppose a Chinook salmon needs to jump a waterfall that is 1.50 m high. If the fish starts from a distance 1.2

Question

Suppose a Chinook salmon needs to jump a waterfall that is 1.50 m high. If the fish starts from a distance 1.20 m from the base of the ledge over which the waterfall flows, find the x- and y-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory.
v0x = ________m/s
v0y = ________m/s

Can the fish make this jump? (Assume that a Chinook salmon can jump out of the water with a speed of 6.26 m/s.)
yes or no?

Answer 1

The fish needs to jump to a height of 1.5 m
y(t)=v0*sin(th)*t-.5*g*t^2
at the moment that vy(t)=0
vy(t)=v0*sin(th)-g*t
so
t=v0*sin(th)/g
and
1.5=.5*v0^2*sin^2(th)/g
or v0^2*sin^2(th)=3*g

The horizontal distance is 1.2 m, so
1.2=v0*cos(th)*v0*sin(th)/g
1.2*g=v0^2*cos(th)*sin(th)
solve for th and v0

v0^2*sin^2(th)=3*g

dividing the two equations one by the other
3/1.2=tan(th)
th=68.2 degrees
v0=5.8 m/s

btw:
v0x=5.8*cos(68.2)
v0y=5.8*sin(68.2)

the salmon will clear the jump since v0=5.8 is the minimum to reach the ledge at an angle of 68.2 degrees (this assumes that the salmon jumps with a ground speed of 6.26 m/s).

j