Hey There!
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Answer:
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DATA:
Angle of projection = θ(theta) = [tex]30^0[/tex]
Initial Velocity = [tex]V_0[/tex] = [tex]2x10^3[/tex]
Acceleration due to gravity = g = 9.8 m/s^2
Vertical Velocity = [tex]V_Y[/tex] = ?
Horizontal Velocity = [tex]V_X[/tex] = ?
Range of the Shell = R = ?
Maximum Height = H = ?
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SOLUTION:
Vertical Velocity is given by,
[tex]V_Y = V_0Cos[/tex]θ
[tex]V_Y = (2x10^3)xCos(30)[/tex]
[tex]V_Y = (2x10)^3x(0.866)\\\\V_Y = 1732.05 \frac{m}{s}[/tex]
Horizontal Velocity is given by,
[tex]V_X = V_0Sin[/tex]θ
[tex]V_X = (2x10^3)xSin(30)\\\\V_X = (2000)x(0.5)\\\\V_X = 1000\frac{m}{s}[/tex]
Range is given by,
R = [tex]\frac{V_0^2}{g}[/tex] Sin2θ
[tex]R = \frac{(2x10^3)^2}{10} x Sin(60)\\\\R = \frac{4x10^6}{10} x 0.866\\\\R= (4x10^5) x 0.866\\\\R = 346410.16 m[/tex]
Horizontal Velocity is given by,
[tex]H = \frac{V_0^2Sin^2(theta)}{2g}\\\\\\H= \frac{(2x10^3)^2xSin(30)^2}{2x10}\\\\\\H= \frac{(4x10^6)x(0.25)}{20} \\\\H = 50000 m[/tex] _____________________________________
Best Regards,
'Borz'
