contestada

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.71 s, and for the (e) horizontal and (f) vertical components at t = 5.44 s. Assume that the catapult is positioned on a plain horizontal ground.

Respuesta :

Answer:

Part a)

[tex]x = 15.76 m[/tex]

Part b)

[tex]y = 7.94 m[/tex]

Part c)

[tex]x = 26.16 m[/tex]

Part d)

[tex]y = 7.49 m[/tex]

Part e)

[tex]x = 83.23 m[/tex]

Part f)

[tex]y = -75.6 m[/tex]

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

[tex]v_x = 19.9 cos39.9[/tex]

[tex]v_x = 15.3 m/s[/tex]

similarly we have

[tex]v_y = 19.9 sin39.9[/tex]

[tex]v_y = 12.76 m/s[/tex]

Part a)

Horizontal displacement in 1.03 s

[tex]x = v_x t[/tex]

[tex]x = (15.3)(1.03)[/tex]

[tex]x = 15.76 m[/tex]

Part b)

Vertical direction we have

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

[tex]y = (12.76)(1.03) - 4.9(1.03)^2[/tex]

[tex]y = 7.94 m[/tex]

Part c)

Horizontal displacement in 1.71 s

[tex]x = v_x t[/tex]

[tex]x = (15.3)(1.71)[/tex]

[tex]x = 26.16 m[/tex]

Part d)

Vertical direction we have

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

[tex]y = (12.76)(1.71) - 4.9(1.71)^2[/tex]

[tex]y = 7.49 m[/tex]

Part e)

Horizontal displacement in 5.44 s

[tex]x = v_x t[/tex]

[tex]x = (15.3)(5.44)[/tex]

[tex]x = 83.23 m[/tex]

Part f)

Vertical direction we have

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

[tex]y = (12.76)(5.44) - 4.9(5.44)^2[/tex]

[tex]y = -75.6 m[/tex]

Otras preguntas