# Computing projectile's maximum height and range.

• MHB
• WMDhamnekar
In summary, the answer is that the level of this course is dependent on your math sophistication. Different methods can be used depending on the level of sophistication. This is a differential equations problem because the acceleration is the second derivative of position. This is also a trajectory problem, so the downward acceleration is gravity. We are also told that there is a "constant horizontal acceleration, g/4, due to the wind."
WMDhamnekar
MHB
Hi,

Here is the question.

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.

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.

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 ...

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

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

Country Boy said:
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,

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 anyone explain me this answer in simple terms?

Last edited:
skeeter said:
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:
$\Delta y = v_0 \sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2$

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 ...

Dhamnekar Winod said:
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?

Country Boy said:
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.

## 1. What is a projectile?

A projectile is any object that is launched into the air and moves along a curved path under the influence of gravity.

## 2. How is the maximum height of a projectile calculated?

The maximum height of a projectile can be calculated using the formula h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 3. What factors affect the maximum height of a projectile?

The maximum height of a projectile is affected by the initial velocity, angle of launch, and the acceleration due to gravity.

## 4. How is the range of a projectile calculated?

The range of a projectile can be calculated using the formula R = (v2sin2θ)/g, where R is the range, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 5. What are some real-world applications of computing a projectile's maximum height and range?

Computing a projectile's maximum height and range is important in fields such as physics, engineering, and sports. It can be used to predict the trajectory of a projectile, such as a ball being thrown or a rocket being launched, and to optimize the design of structures and equipment that involve projectiles, such as bridges and catapults. In sports, knowing the maximum height and range of a projectile can help athletes improve their performance in activities like throwing a ball or shooting a basketball.

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