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Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.710 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.115 m above the floor. (Assume the center of the hoop is at the peak height of the jump. Assume that +x axis is in the direction of the hoop from the cabinet and +y axis is up. Assume g = 9.81 m/s2.)
(a) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.569 m from the cabinet? (Express your answer in vector form. Give your answer to at least two decimal places for each vector component.)
(b) If Kit lands on a bed at a horizontal distance of 3.591 m from the cabinet, how high above the ground is the bed?

Respuesta :

Answer:

Part a)

[tex]v = 6.013 m/s[/tex]

Part b)

[tex]H = 0.8 m[/tex]

Explanation:

Part a)

Maximum height of the center of the loop is

[tex]y_2 = 3.115 m[/tex]

initial position of the cat is

[tex]y_1 = 1.710 m[/tex]

now when cat will reach the maximum height its velocity will become zero

so we have

[tex]v_f^2 - v_i^2 = 2 a(y_2 - y_1)[/tex]

[tex]0 - v_i^2 = 2(-9.81)(3.115 - 1.710)[/tex]

[tex]v_i = 5.25 m/s[/tex]

now the time taken by the cat to reach this height is given as

[tex]v_f - v_i = at[/tex]

[tex]0 - 5.25 = (-9.81)t[/tex]

[tex]t = 0.535 s[/tex]

now the horizontal speed of the cat is given as

[tex]v_x = \frac{x}{t}[/tex]

[tex]v_x = \frac{1.569}{0.535} [/tex]

[tex]v_x = 2.93 m/s[/tex]

so the net initial speed of the cat is given as

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

[tex]v = \sqrt{2.93^2 + 5.25^2}[/tex]

[tex]v = 6.013 m/s[/tex]

Part b)

horizontal distance of the bed is given as

[tex]x = 3.591 m[/tex]

horizontal speed of the cat is given as

[tex]v_x = 2.93 m/s[/tex]

so the time taken by the cat to reach this position is

[tex]t = \frac{x}{v_x}[/tex]

[tex]t = \frac{3.591}{2.93} = 1.23 s[/tex]

now in the same time displacement in y direction is given as

[tex]\Delta y = v_i t + \frac{1}{2}at^2[/tex]

[tex]\Delta y = (5.25)(1.23) - \frac{1}{2}(9.81)(1.23^2)[/tex]

[tex]\Delta y = -0.91 m[/tex]

so height of the bed is given as

[tex]H = 1.710 - 0.91[/tex]

[tex]H = 0.8 m[/tex]

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