Question

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish travels 7.9 m horizontally before hitting the water.

What is the velocity of the eagle? Round your answer to the nearest tenth.

Answer 1

The answer is 7.1 I just took the test

Answer 2

Answer:

Explanation:

We can use one of the SUVAT equations like:

$s = ut + \frac{1}{2} a^{2}$

where s = displacement, u = initial velocity, a = the acceleration and t = time.

To get the velocity of the eagle, we have to look at motion of the fish first, using the SUVAT equation form above, therefore:  

Vertical motion, which is going downwards

displacement = 6.1 m

Initial velocity = 0 m/s ,

acceleration = 9.8 $m/s^{2}$

So, we need to find the time.

$6.1 = 0t + \frac{1}{2} 9.8t^{2}\\6.1 = 4.9t^{2} \\t^{2}= \frac{6.1}{4.9}\\t = \sqrt{1.2448}\\t = 1.116 s$

Now with the time, we can look at the horizontal motion of the fish, where the variable of the initial velocity, u, is what we need to find, which is also the velocity of the eagle:

Acceleration = 0 $m/s^{2}$

Displacement = 7.9 m ,

time = 1.116 seconds we need to find Initial velocity (u).

$7.9 = v*1.116 + \frac{1}{2} 0*1.116^{2}\\7.9 = 1.116v \\v= \frac{7.9}{1.116} \\ v= 7.1 m/s$

       

Therefore, velocity of the eagle = 7.1 m/s