Question #9243f

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Question #9243f

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Answer 1 · 2017-09-07T14:20:13

Consider it being thrown with an angle $θ$ $theta$ with the horizontal.

Then the velocity has two components $u_{x} = U cos θ$ $u_x = Ucos theta$ and $u_{y} = U sin θ$ $u_y = USin theta$ .

The horizontal acceleration is $a_{x} = 0$ $a_x = 0$ while that in vertical direction is $a_{y} = - g$ $a_y = -g$

From an equation from kinematics,

$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$ $y = u_yt + 1/2a_yt^2$
$\Rightarrow y = U sin θ t - \frac{1}{2} g t^{2}$ $implies y = USin thetat - 1/2 g t^2$ is distance travelled by particle in y direction (vertical direction) in time $t$ $t$ .

Also, $x = u_{x} t + \frac{1}{2} a_{x} t^{2}$ $x = u_xt + 1/2a_xt^2$
$\Rightarrow x = U cos θ t$ $implies x = UCos thetat$

Combining $x$ $x$ and $y$ $y$ ,

$y = x tan θ - \frac{1}{2} \frac{g x^{2}}{{cos}^{2} θ}$ $y = xtan theta - 1/2(gx^2)/Cos^2 theta$

But for $θ$ $theta$ constant, $tan θ = a$ $tan theta = a$ and $b = - \frac{g}{2 {cos}^{2} θ}$ $b = -g/(2Cos^2 theta)$ ,

We get,
$y = a x + b x^{2}$ $y = ax + bx^2$

Which shows that the path is parabolic.

The equations of motion are,

$F_{x} = m a_{x} = 0$ $F_x = ma_x = 0$
$F_{y} = m a_{y} = - m g$ $F_y = ma_y = -mg$

If however, the body is thrown vertically up, $θ = \frac{π}{2}$ $theta = pi/2$

And thats when the $x$ $x$ component of motion ceases the exist and, motion can be described as, by acceleration $a_{y} = - g$ $a_y = -g$ .

Then, $y = U t - \frac{1}{2} g t^{2}$ $y = Ut - 1/2 g t^2$ describes the vertical distance travelled.

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Question #9243f

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