Question

Salmon often jump waterfalls to reach their
breeding grounds.Starting downstream, 1.57 m away from a
waterfall 0.339 m in height, at what minimum
speed must a salmon jumping at an angle of
43.2◦leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s
2.

Answer 1

Answer:

Using the range formula R = v^2 sin 2 theta / g

or v^2 = R * g / sin 86.4

v^2 = 3.14 m * 9.81 m/s2 / .998

v^2 = 30.9 m^2 / s^2

v = 5.56 m/s

This hasn't  really proved the question - this would give

vy = 5.56 * sin 43.2 = 3.81 m/s

vx = 5.56 * cos 43.2. = 4.05 m/s

t = 1.57 / 4.05 = .387 sec to reach the waterfall

h = 3.81 * .387 - 4.9 (.387)^2 = .74 m     well above the height of the falls

There seems another way to do this

vy / vx = tan 43.2       vy = .939 vx

h = vy t - 1/2 g t^2       and t = 1.57 / vx

h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2

Solving for vx I get vx = 3.26 m/s    vy = 3.06 m/s  v = 4.47 m/s