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A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90 m/s at an angle of 23.0° below the horizontal. It strikes the ground 5.00 s later.(a) How far horizontally from the base of the building does the ball strike the ground?1 m(b) Find the height from which the ball was thrown.2Your response differs from the correct answer by more than 10%. Double check your calculations. m(c) How long does it take the ball to reach a point 10.0 m below the level of launching?3 s

Respuesta :

Answer:

Part a)

[tex]\Delta x = 36.35 m[/tex]

Part b)

height will be 107.2 m

Part c)

[tex]t = 1.78 s[/tex]

Explanation:

Initial velocity of the ball is given as

v = 7.90 m/s

angle of projection of the ball

[tex]\theta = 23^o[/tex]

now the two components of velocity of ball is given as

[tex]v_x = 7.90 cos23 = 7.27 m/s[/tex]

[tex]v_y = 7.90 sin23 = 3.09 m/s[/tex]

Part a)

Since ball strike the ground after t = 5 s

so the distance moved by the ball in horizontal direction is given as

[tex]\Delta x = v_x \times t[/tex]

[tex]\Delta x = 7.27 (5)[/tex]

[tex]\Delta x = 36.35 m[/tex]

Part b)

Now in order to find the height of the ball we can find the vertical displacement of the ball

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

[tex]\Delta y = (3.09)(5) - \frac{1}{2}(9.81)(5^2)[/tex]

[tex]\Delta y = -107.2 m[/tex]

So height will be 107.2 m

Part c)

when ball reaches a point 10 m below the level of launching then the displacement of the ball is given as

[tex]\Delta y = -10 m[/tex]

so we will have

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

[tex]-10 = 3.09 t - \frac{1}{2}(9.81)t^2[/tex]

so we will have

[tex]t = 1.78 s[/tex]

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