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The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 4.9 m/s. The distance to the ground from that point is 6.4 m. (a) How long is the ball in the air? s (b) How far from the base of the house does it land? m (c) What is its velocity just before landing? (Let upward be the positive y-direction.) x-component m/s y-component m/s

Respuesta :

Answer:

[tex]t = 0.93 s[/tex]

Part b)

[tex]d = 3.98 m[/tex]

Part c)

[tex]v_x = 4.28 m/s[/tex]

[tex]v_y = -11.49 m/s[/tex]

Explanation:

The two components of the velocity of the ball is given as

[tex]v_x = 4.9 cos29[/tex]

[tex]v_x = 4.28 m/s[/tex]

[tex]v_y = 4.9 sin29[/tex]

[tex]v_y = 2.37 m/s[/tex]

Part a)

now we know that the displacement in y direction is given as

[tex]\Delta y = 6.4 m[/tex]

so we have

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

[tex]6.4 = 2.37 t + 4.9 t^2[/tex]

[tex]t = 0.93 s[/tex]

Part b)

Distance of the ball in x direction of the motion is given as

[tex]d = v_x t[/tex]

[tex]d = 4.28 \times 0.93[/tex]

[tex]d = 3.98 m[/tex]

Part c)

In x direction the velocity will remain the same always

[tex]v_x = 4.28 m/s[/tex]

while in Y direction we can use kinematics

[tex]v_y = v_{oy} + at[/tex]

[tex]v_y = -2.37 - 9.81(0.93)[/tex]

[tex]v_y = -11.49 m/s[/tex]

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