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aleksandrvk [35]
3 years ago
12

An eagle is flying horizontally at a speed of 3m/s when the fish in her talons wiggles loose and falls into the lake 10m below.

Calculate the velocity of the fish relative to the water when it hits the water.
Physics
1 answer:
Kamila [148]3 years ago
7 0

Answer:

Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

Explanation:

Vertical motion of fish:

 Initial speed, u = 0

 Acceleration, a = 9.81 m/s²

 Displacement, s = 10 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 10 = 196.2

    v = 14 m/s

 Final vertical speed = 14 m/s

 Final horizontal speed = initial horizontal speed = 3 m/s

 Final velocity = 3 i - 14 j m/s

 Magnitude

     v=\sqrt{3^2+(-14)^2}=14.32m/s

 Direction

      \theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0

 Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

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4 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 +  v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0
4 years ago
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