Respuesta :
Answer:
[tex]v_o = 7.76 m/s[/tex]
Explanation:
As we projected the balloon at speed vo at an angle of 29 degree
so the two component of velocity is given as
[tex]v_x = v_ocos29 = 0.875 v_o[/tex]
[tex]v_y = v_o sin29 = 0.485 v_o[/tex]
now we know that in x direction we have
[tex]d = v_x t[/tex]
[tex]7.5 = 0.875 v_o t[/tex]
[tex]v_o t = 8.57[/tex]
in y direction we have
[tex]- 1.82 = (0.485 v_o) t - \frac{1}{2}gt^2[/tex]
[tex]-1.82 = 0.485(8.57) - 4.9 t^2[/tex]
[tex]t = 1.1 s[/tex]
now we have
[tex]v_o = 7.76 m/s[/tex]
The magnitude of the balloon's initial velocity is 7.80 m/s
The given parameter
- height from which the balloon was thrown, h = 1.82 m
- the initial velocity of the balloon, = v₀
- the angle of projection, θ = 29⁰
- the horizontal distance of the balloon, x = 7.5 m
To find:
- the magnitude of the balloon's initial velocity
The vertical component of the velocity is calculated as;
[tex]v_y = v_0\times sin(\theta)\\\\v_y = v_0 \times sin(29)\\\\v_y = 0.4848v_0[/tex]
The horizontal component of the velocity is given as;
[tex]v_x = v_0\times cos(\theta)\\\\v_x = v_0 \times cos(29)\\\\v_x = 0.8746v_0[/tex]
The horizontal distance or range of the projectile is given as;
[tex]X = v_xt \\\\7.5 = 0.8746v_0t\\\\v_0t = \frac{7.5}{0.8746} \\\\v_ot = 8.575 \[/tex]
The change in vertical height of the balloon is given as;
[tex]\Delta h = v_yt - \frac{1}{2}gt^2\\\\h_2-h_1 = v_yt - \frac{1}{2}gt^2[/tex]
assume the balloon falls to zero position
[tex]0-1.82 = v_yt - \frac{1}{2} gt^2\\\\-1.82 = v_yt - \frac{1}{2} gt^2\\\\-1.82 = 0.4848v_0t - \frac{1}{2} (9.8)t^2\\\\-1.82 = 0.4848(8.575) - 4.9t^2\\\\4.9t^2 = 4.157 + 1.82\\\\4.9t^2 = 5.977\\\\t^2 = \frac{5.977}{4.9} \\\\t^2 = 1.2198\\\\t = \sqrt{1.2198} \\\\t = 1.1 \ s[/tex]
The magnitude of initial velocity of the balloon is calculated as;
[tex]v_0t = 8.575\\\\1.1v_0 = 8.575\\\\v_0 = \frac{8.575}{1.1} \\\\v_0 = 7.80 \ m/s[/tex]
Thus, the magnitude of the balloon's initial velocity is 7.80 m/s
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