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A squirrel on a limb near the top of a tree loses its grip on a nut, so that the nut slips away horizontally at a speed of 1.5m/s. If the nut lands at a horizontal distance of 7.2m, how high above the ground is the squirrel?

Respuesta :

okpalawalter8

Answer:

The height of the squirrel  [tex]h =113 \ m[/tex]

Explanation:

From the question we are told that  

    The horizontally speed is [tex]v = 1.5 m/s[/tex]

    The horizontal  distance [tex]d = 7.2 \ m[/tex]

Generally the the time taken is mathematically represented

      [tex]t = \frac{d}{v}[/tex]

=>   [tex]t = \frac{7.2}{1.5}[/tex]

=>   [tex]t = 4.8 \ s[/tex]

Generally from kinematic equation we have that

     [tex]h = ut + \frac{1}{2} gt^2[/tex]

Here u  is the initial velocity of nut in the vertical and the value is 0 m/s

     [tex]h = 0 * 4.8 + \frac{1}{2} * 9.8 * (4.8)^2[/tex]

=>  [tex]h =113 \ m[/tex]

boffeemadrid

The height at which the squirrel is present is required.

The height at which the squirrel is present is 113.01 m.

[tex]u_x[/tex] = Initial horizontal velocity = 1.5 m/s

[tex]x[/tex] = Horizontal distance = 7.2 m

[tex]a_y[/tex] = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]u_y[/tex] = Initial vertical velocity = 0

Time taken is

[tex]t=\dfrac{x}{u_x}\\\Rightarrow t=\dfrac{7.2}{1.5}\\\Rightarrow t=4.8\ \text{s}[/tex]

Vertical distance is given by

[tex]y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=0\times 4.8+\dfrac{1}{2}\times 9.81\times 4.8^2\\\Rightarrow y=113.01\ \text{m}[/tex]

The height at which the squirrel is present is 113.01 m.

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https://brainly.com/question/15224155?referrer=searchResults

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