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Computing projectile's maximum height and range.

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hi,

Here is the question.

1608891252345.png

Answer given is d. But i don't understand how is that computed? I am working on this question. Meanwhile any member knowing the correct answer may help me in finding out correct answer.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
756
What level course is this for? There are several different methods that can be used depending on your "math sophistication". I would treat this as a differential equations problem because the problem gives the acceleration and acceleration is the second derivative of position. This is a trajectory problem so the downward acceleration is that of gravity, -g. We are also told that there is a "constant horizontal acceleration, g/4, due to the wind.

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.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
Assuming the projectile is launched and lands at the same height ...

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 ...

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

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

Dhamnekar Winod

Active member
Nov 17, 2018
149
What level course is this for? There are several different methods that can be used depending on your "math sophistication". I would treat this as a differential equations problem because the problem gives the acceleration and acceleration is the second derivative of position. This is a trajectory problem so the downward acceleration is that of gravity, -g. We are also told that there is a "constant horizontal acceleration, g/4, due to the wind.

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.
Hello,
Thanks for your guidance.

1609052578930.png

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.

1609053618653.png

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

Dhamnekar Winod

Active member
Nov 17, 2018
149
Assuming the projectile is launched and lands at the same height ...

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 ...

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

$\Delta x = R + \dfrac{g}{8} \cdot \dfrac{4v_{0y}^2}{g^2} = R + \dfrac{v_{0y}^2}{2g} = R+H$
Hello,
You have not mentioned about any change occurred in height of projectile prior to acceleration and after acceleration. Would you explain me how $R=V_{0x} \cdot t_{total}=\frac{(V_0)^2}{g}\cdot \sin{(2\cdot \theta_0)}?$
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
$\Delta y = v_0 \sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2$

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 ...
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
756
Hello,
You have not mentioned about any change occurred in height of projectile prior to acceleration and after acceleration. Would you explain me how $R=V_{0x} \cdot t_{total}=\frac{(V_0)^2}{g}\cdot \sin{(2\cdot \theta_0)}?$
What do you mean by "prior to acceleration" and "after acceleration"? Isn't there always acceleration?
 

Dhamnekar Winod

Active member
Nov 17, 2018
149
What do you mean by "prior to acceleration" and "after acceleration"? Isn't there always acceleration?
Hello,
Oh, I forgot that the acceleration is in horizontal direction. So there will not be any change in maximum height.