# Max Horizontal Displacement of Projectile and its Velocity at T

• MHB
• Theia
In summary, the equations x = v_0\cos \alpha _0 t and y = y_0 + v_0 \sin \alpha _0 t - \tfrac{1}{2} gt^2 describe the position of a projectile in terms of time, initial speed, initial angle, and acceleration due to gravity. To find the maximum horizontal displacement of the projectile, use the formula x_{max} = \dfrac{v_0^2}{g\tan{\alpha_0}}. The angle between the initial velocity vector and the velocity vector at the time the projectile hits the ground (T) is \pi/2. Additionally, the value of \alpha_0 that yields a maximum horizontal range
Theia
Let $$\displaystyle x = v_0\cos \alpha _0 t$$ and $$\displaystyle y = y_0 + v_0 \sin \alpha _0 t - \tfrac{1}{2} gt^2$$, where

• $$\displaystyle v_0$$ is speed at time $$\displaystyle t = 0$$,
• $$\displaystyle \alpha _0$$ is the angle between positive $$\displaystyle x$$-axis and initial velocity vector ($$\displaystyle \alpha _0 \in (0, \pi/2)$$),
• $$\displaystyle t$$ time in seconds,
• $$\displaystyle y_0 >0$$ the $$\displaystyle y$$ coordinate at time $$\displaystyle t=0$$,
• $$\displaystyle g$$ acceleration due the gravity.

Let $$\displaystyle T$$ be the time when the projectile hits positive $$\displaystyle x$$-axis (i.e. the ground). Find the maximum horizontal displacement of the projectile and show that angle between initial velocity vector and velocity vector at time $$\displaystyle T$$ is $$\displaystyle \pi/2$$.

$x_{max} = \dfrac{v_0^2}{g\tan{\alpha_0}}$

(bonus info that can be determined in working this problem) ... the value of $\alpha_0$ that yields a maximum horizontal range is $\alpha_0 = \arctan\left(\dfrac{v_0}{\sqrt{v_0^2 + 2gy_0}}\right)$

let $a = \alpha_0$, initial launch angle (too lazy to type out the Latex)

$b$ = final impact angle at time $T$$T=\dfrac{x}{v_0\cos{a}}$ substitute this expression for $T$ into the equation for $y$. Note $y(T)=0$

$0=y_0 + x\tan{a} - \dfrac{g}{2v_0^2} \cdot x^2\sec^2{a}$

derivative w/respect to $a$ ...

$0 = x\sec^2{a} + \tan{a} \cdot \dfrac{dx}{da} - \dfrac{g}{v_0^2} \left(x^2 \sec^2{a}\tan{a}+x\sec^2{a} \cdot \dfrac{dx}{da} \right)$

when $x$ is a maximum, $\dfrac{dx}{da} = 0$ ...

$0 = x\sec^2{a} - \dfrac{g}{v_0^2} \left(x^2 \sec^2{a}\tan{a}\right)$

$0 = x\sec^2{a}\bigg[1 - \dfrac{g}{v_0^2} \left(x \tan{a}\right) \bigg]$

$x\sec^2{a} \ne 0 \implies x = \dfrac{v_0^2}{g\tan{a}} \implies T = \dfrac{v_0}{g\sin{a}}$

$v_{fx} = v_0\cos{a}$

$v_{fy} = v_0\sin{a} - gT = -v_0 \cdot \dfrac{\cos^2{a}}{\sin{a}}$

$\tan{b} = \dfrac{v_{fy}}{v_{fx}} = -\cot{a}$

$\tan{b} \cdot \tan{a} = -\cot{a} \cdot \tan{a} = -1 \implies a \perp b$

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Thank you for your solution! ^^

## 1. What is the formula for calculating the maximum horizontal displacement of a projectile?

The formula for calculating the maximum horizontal displacement of a projectile is d = v0cosθ × t, where d is the maximum horizontal displacement, v0 is the initial velocity, θ is the angle of launch, and t is the time of flight.

## 2. How does the angle of launch affect the maximum horizontal displacement of a projectile?

The angle of launch directly affects the maximum horizontal displacement of a projectile. The maximum horizontal displacement will be greatest when the angle of launch is 45 degrees, and decreases as the angle of launch deviates from 45 degrees. This is because at 45 degrees, the initial velocity is split evenly between the horizontal and vertical components, resulting in the greatest distance travelled in the horizontal direction.

## 3. What factors can affect the velocity of a projectile at a given time?

The velocity of a projectile at a given time can be affected by several factors including the initial velocity, angle of launch, air resistance, and gravitational force. Additionally, any external forces acting on the projectile, such as wind or friction, can also affect its velocity at a given time.

## 4. How does air resistance impact the velocity of a projectile?

Air resistance can significantly impact the velocity of a projectile, especially over longer distances. As the projectile travels through the air, it experiences a force in the opposite direction of its motion due to air resistance. This force can decrease the velocity of the projectile and cause it to deviate from its expected path.

## 5. Is the maximum horizontal displacement of a projectile always equal to the range?

No, the maximum horizontal displacement of a projectile is not always equal to the range. The range refers to the total distance travelled by the projectile, including any vertical displacement. The maximum horizontal displacement only considers the distance travelled in the horizontal direction, so it may be less than the range if the projectile has a significant vertical component to its motion.

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