Understanding Projectile Motion and Finding the Angle of Velocity Vector

In this article, we will explore the principles of projectile motion and how to solve a specific problem related to the range and maximum height of a projectile. We will delve into key concepts such as the range (R) and the maximum height (H), and analyze the velocity vector at a specific horizontal distance from the point of projection.

Key Concepts

Prior to solving the problem, it's important to understand the basic concepts involved in projectile motion:

Range (R)

The horizontal distance traveled by the projectile when it returns to the same vertical level from which it was launched.

Maximum Height (H)

The highest point reached by the projectile.

Given Condition

The problem states that the range R is twice the maximum height H. This relationship can be mathematically expressed as:

R 2H

Projectile Motion Equations

For a projectile launched with an initial speed v_0 at an angle u03B8:

The maximum height H is given by:

H frac{v_0^2 sin^2 u03B8}{2g}

The range R is given by:

R frac{v_0^2 sin 2u03B8}{g}

Relating Range and Height

Using the relation R 2H, we can derive the angle u03B8. This involves simplifying the given equations and using trigonometric identities:

frac{v_0^2 sin 2u03B8}{g} 2 left(frac{v_0^2 sin^2 u03B8}{2g}right)

Simplifying this yields:

sin 2u03B8 2 sin^2 u03B8

Using the double-angle identity for sine:

sin 2u03B8 2 sin u03B8 cos u03B8

Setting the equations equal and solving:

2 sin u03B8 cos u03B8 2 sin^2 u03B8

Dividing both sides by 2 and rearranging:

cos u03B8 sin u03B8 quad Rightarrow quad tan u03B8 1 quad Rightarrow quad u03B8 45^circ

Time to Reach R/4

Now, we need to determine the angle made by the velocity vector with the horizontal at a horizontal distanceR/4.

Total Time of Flight

The total time of flight T for a projectile is given by:

T frac{2 v_0 sin u03B8}{g}

Range R

When u03B8 45^circ, the range R is:

R frac{v_0^2}{g}

Time to Reach R/4

The time to reach R/4 is:

t frac{T}{4} frac{1}{4} cdot frac{2 v_0 sin 45^circ}{g} frac{v_0 cdot frac{sqrt{2}}{2}}{2g} frac{v_0 sqrt{2}}{4g}

Vertical and Horizontal Components of Velocity

The horizontal velocity v_x is constant:

v_x v_0 cos 45^circ frac{v_0 sqrt{2}}{2}

The vertical velocity v_y at time t is:

v_y v_0 sin 45^circ - gt frac{v_0 sqrt{2}}{2} - g cdot frac{v_0 sqrt{2}}{4g} frac{v_0 sqrt{2}}{2} - frac{v_0 sqrt{2}}{4} frac{v_0 sqrt{2}}{4}

Angle of Velocity Vector

The angle u03B5 made by the velocity vector with the horizontal is given by:

tan u03B5 frac{v_y}{v_x} frac{frac{v_0 sqrt{2}}{4}}{frac{v_0 sqrt{2}}{2}} frac{1}{2}

Thus, the angle u03B5 is:

u03B5 tan^{-1}left(frac{1}{2}right) approx 26.57^circ

Final Answer

The angle made by the velocity vector with the horizontal at a horizontal distance R/4 from the point of projection is:

u03B5 tan^{-1}left(frac{1}{2}right) approx 26.57^circ

This concludes the detailed analysis of the problem, illustrating how the principles of projectile motion are applied to solve complex scenarios involving range and maximum height.