Question

Salmon often jump waterfalls to reach their breeding grounds.

Answer 1

Answer:

5.83 m/s

Explanation:

This is a projectile motion problem.

Let's call

$V_0$ the initial velocity

In x-coordinate:

$V_{0x} = V_0 \times cos \alpha$

where α is 44.7°

In y-coordinate:

$V_{0y} = V_0 \times sin \alpha$

In x-coordinate the displacement is:

$x = V_0 \times cos \alpha \times t$

where t is time. Isolating the initial speed and replacing with x = 2.9 (the distance travelled in x direction) :

$V_0 = \frac{2.9}{cos \alpha \times t}$

In y-coordinate the displacement is:

$y = V_0 \times sin \alpha \times t - 1/2 \times g \times t^2$

Replacing with the initial velocity, y = 0.436, alpha = 44.7° and g = 9.81:

$0.436 = \frac{2.9}{cos \alpha \times t} \times sin \alpha \times t - 1/2 \times g \times t^2$

$0.436 =2.9 \times tan 44.7 - 1/2 \times 9.81 \times t^2$

$t^2 = \frac{2.9 \times tan 44.7 - 0.436}{ 1/2 \times 9.81}$

$t = \sqrt{0.495}$

$t = 0.7$

Replacing this value in the previous initial velocity equation:

$V_0 = \frac{2.9}{cos 44.7 \times 0.7}$

$V_0 = 5.83 \; m/s$

Answer 2

Answer:

5.79 m/s

Explanation:

R=2.9 m

H=0.436 m

∅=44.7°

For Horizontal Motion

aₓ=0      uₓ=ucos∅

x=uₓt+$\frac{1axt^{2} }{2}$

R=ucos∅*t+0

2.9=ucos(38.3)*t

t=$\frac{4.079}{u}$

For Vertical Motion

Y=Uy*t+$\frac{1ayt^{2} }{2}$

By putting value

0.436=usin∅($\frac{4.079}{u}$)-($\frac{1}{2}g *(\frac{4.079}{u} )^{2}$)

0.436=sin(44.7)($\frac{4.079}$)-($\frac{1}{2}g *(\frac{4.079}{u} )^{2}$)

0.436=2.869-$\frac{81.61}{u^{2} }$

u²=33.54

u=5.79 m/s