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#### Dhamnekar Winod

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- Nov 17, 2018

- 162

- Thread starter Dhamnekar Winod
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- Thread starter
- #1

- Nov 17, 2018

- 162

- Jan 30, 2018

- 799

Take x to be the horizontal distance from the starting position and take y to be the horizontal distance from the starting point.

First do the calculation ignoring the wind so that \(\displaystyle \frac{d^2x}{dt^2}= 0\) and \(\displaystyle \frac{d^2y}{dt^2}= -g\). Taking the range and height to be "R" and "H" what must be the initial velocity vector have been?

Then we have \(\displaystyle \frac{d^2x}{dt^2}= g/4\) and \(\displaystyle \frac{d^2y}{dt^2}= -g\). Integrate each of those to get the velocity vector and then integrate again to get the position. Each integration will give "constants of integration". Determine their values by using the facts that the initial position is (0, 0) and the known initial velocity.

- Mar 1, 2012

- 980

Motion in the vertical & horizontal directions are independent. Since acceleration in the vertical is unchanged, the maximum height is also unchanged.

Further, at the top of the projectile’s trajectory, the y-component of velocity is zero ...

$0 = v_{fy} = v_{0y} - gt \implies t_{top} = \dfrac{v_{0y}}{g}$

therefore, $H = v_{0y} \cdot t_{top} - \dfrac{g}{2} \cdot t_{top}^2$

substituting for $t_{top}$ yields $H = \dfrac{v_{0y}^2}{2g}$

with no acceleration in the horizontal direction, $R=v_x \cdot t_{total}$

note $t_{total} = 2t_{top}$

with acceleration in the x-direction ...

$\Delta x = v_{0x} \cdot t_{total} + \dfrac{1}{2}a_x \cdot t_{total}^2$

$\Delta x = R + \dfrac{g}{8} \cdot \dfrac{4v_{0y}^2}{g^2} = R + \dfrac{v_{0y}^2}{2g} = R+H$

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- #4

- Nov 17, 2018

- 162

Hello,

Take x to be the horizontal distance from the starting position and take y to be the horizontal distance from the starting point.

First do the calculation ignoring the wind so that \(\displaystyle \frac{d^2x}{dt^2}= 0\) and \(\displaystyle \frac{d^2y}{dt^2}= -g\). Taking the range and height to be "R" and "H" what must be the initial velocity vector have been?

Then we have \(\displaystyle \frac{d^2x}{dt^2}= g/4\) and \(\displaystyle \frac{d^2y}{dt^2}= -g\). Integrate each of those to get the velocity vector and then integrate again to get the position. Each integration will give "constants of integration". Determine their values by using the facts that the initial position is (0, 0) and the known initial velocity.

Thanks for your guidance.

Above diagram depicts projectile motion. Let us assume following variables

1) $X_0$ = Initial X-position

2)X=Final X-position $X=X_0 + V_0\cdot \cos{\theta_0}\cdot t$

3)$Y_0$= Initial Y-position

4)Y= Final Y-position $Y=Y_0 + V_0\cdot \sin{\theta_0}\cdot t -\frac12 \cdot g \cdot t^2$

5)$\theta_0$= initial angle

6)$V_0$=Initial velocity

7)$V_x$=X-velocity $V_x= V_0\cdot \cos{\theta_0}$

8)$V_y$=Y-velocity $V_y=V_0\cdot \sin{\theta_0}-g\cdot t$

9)t= time

10) R=Horizontal range. $R=\frac{(V_0)^2}{g}\cdot \sin{(2\cdot \theta_0)}$

Now, how we can proceed further to answer this question under this method? Following is the answer, I am provided with.

I don't understand this answer because of illegible handwriting. Would any one explain me this answer in simple terms?

Last edited:

- Thread starter
- #5

- Nov 17, 2018

- 162

Hello,

Motion in the vertical & horizontal directions are independent. Since acceleration in the vertical is unchanged, the maximum height is also unchanged.

Further, at the top of the projectile’s trajectory, the y-component of velocity is zero ...

$0 = v_{fy} = v_{0y} - gt \implies t_{top} = \dfrac{v_{0y}}{g}$

therefore, $H = v_{0y} \cdot t_{top} - \dfrac{g}{2} \cdot t_{top}^2$

substituting for $t_{top}$ yields $H = \dfrac{v_{0y}^2}{2g}$

with no acceleration in the horizontal direction, $R=v_x \cdot t_{total}$

note $t_{total} = 2t_{top}$

with acceleration in the x-direction ...

$\Delta x = v_{0x} \cdot t_{total} + \dfrac{1}{2}a_x \cdot t_{total}^2$

$\Delta x = R + \dfrac{g}{8} \cdot \dfrac{4v_{0y}^2}{g^2} = R + \dfrac{v_{0y}^2}{2g} = R+H$

You have not mentioned about any change occurred in height of projectile prior to acceleration and after acceleration. Would you explain me how $R=V_{0x} \cdot t_{total}=\frac{(V_0)^2}{g}\cdot \sin{(2\cdot \theta_0)}?$

Last edited:

- Mar 1, 2012

- 980

the projectile lands at its starting height $\implies \Delta y = 0$

$0 = t\left(v_0 \sin{\theta_0} - \dfrac{1}{2}gt \right) \implies v_0\sin{\theta_0} = \dfrac{1}{2}gt \implies t = \dfrac{2v_0\sin{\theta_0}}{g}$

with no acceleration in the horizontal direction ...

$\Delta x = v_0\cos{\theta_0} \cdot t = v_0\cos{\theta_0} \cdot \dfrac{2v_0 \sin{\theta_0}}{g} = \dfrac{v_0^2 \cdot 2\sin{\theta_0}\cos{\theta_0}}{g}$

now, recall the double angle identity for sine ...

- Jan 30, 2018

- 799

What do you mean by "prior to acceleration" and "after acceleration"? Isn't there always acceleration?Hello,

You have not mentioned about any change occurred in height of projectile prior to acceleration and after acceleration. Would you explain me how $R=V_{0x} \cdot t_{total}=\frac{(V_0)^2}{g}\cdot \sin{(2\cdot \theta_0)}?$

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- #8

- Nov 17, 2018

- 162

Hello,What do you mean by "prior to acceleration" and "after acceleration"? Isn't there always acceleration?

Oh, I forgot that the acceleration is in horizontal direction. So there will not be any change in maximum height.