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From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle of 18.0° below the horizontal. It strikes the ground 4.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0.) xi = yi = (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. vi,x = m/s vi,y = m/s (c) Find the equations for the x- and y- components of the position as functions of time. (Use the following as necessary: y0 and t. Let the variable t be measured in seconds.) x = m y = m (d) How far horizontally from the base of the building does the ball strike the ground? m (e) Find the height from which the ball was thrown. m (f) How long does it take the ball to reach a point 10.0 m below the level of launching? s

Respuesta :

Answer:

Part a)

[tex]y = 88.5 m[/tex]

Part b)

[tex]v_x = 7.7 m/s[/tex]

[tex]v_y = 2.5 m/s[/tex]

Part c)

equation of position in x direction is given as

[tex]x = v_x t[/tex]

equation of position in y direction is given as

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

Part d)

[tex]x = 30.8 m[/tex]

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

[tex]v_x = 8.10 cos18[/tex]

[tex]v_x = 7.7 m/s[/tex]

[tex]v_y = 8.10 sin18[/tex]

[tex]v_y = 2.5 m/s[/tex]

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

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

[tex]y = 2.5(4) + \frac{1}{2}(9.81)(4^2)[/tex]

[tex]y = 88.5 m[/tex]

Part b)

components of the velocity is given as

[tex]v_x = 8.10 cos18[/tex]

[tex]v_x = 7.7 m/s[/tex]

[tex]v_y = 8.10 sin18[/tex]

[tex]v_y = 2.5 m/s[/tex]

Part c)

equation of position in x direction is given as

[tex]x = v_x t[/tex]

equation of position in y direction is given as

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

Part d)

distance where it will strike the floor is given as

[tex]x = v_x t[/tex]

[tex]x = 7.7 \times 4[/tex]

[tex]x = 30.8 m[/tex]

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

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

[tex]10 = 2.5t + \frac{1}{2}(9.81) t^2[/tex]

t = 1.2 s

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