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3 years ago

I need help with these questions :(see image )

Physics

1 answer:

lara31 [8.8K]3 years ago

7 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Angle of projection = θ(theta) = $30^0$

Initial Velocity = $V_0$ = $2x10^3$

Acceleration due to gravity = g = 9.8 m/s^2

Vertical Velocity = $V_Y$ = ?

Horizontal Velocity = $V_X$ = ?

Range of the Shell = R = ?

Maximum Height = H = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

Vertical Velocity is given by,

$V_Y = V_0Cos$ θ

$V_Y = (2x10^3)xCos(30)$

$V_Y = (2x10)^3x(0.866)\\\\V_Y = 1732.05 \frac{m}{s}$

Horizontal Velocity is given by,

$V_X = V_0Sin$ θ

$V_X = (2x10^3)xSin(30)\\\\V_X = (2000)x(0.5)\\\\V_X = 1000\frac{m}{s}$

Range is given by,

R = $\frac{V_0^2}{g}$ Sin2θ

$R = \frac{(2x10^3)^2}{10} x Sin(60)\\\\R = \frac{4x10^6}{10} x 0.866\\\\R= (4x10^5) x 0.866\\\\R = 346410.16 m$

Horizontal Velocity is given by,

<h2> $H = \frac{V_0^2Sin^2(theta)}{2g}\\\\\\H= \frac{(2x10^3)^2xSin(30)^2}{2x10}\\\\\\H= \frac{(4x10^6)x(0.25)}{20} \\\\H = 50000 m$ _____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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