Introduction to Projectile Motion: A 4:1 Riddle

Projectile motion is a fascinating realm of mechanics, where objects are launched into the air with an initial velocity and follow a curved path under the influence of gravity. A common problem in projectile motion is to determine the angle of projection required for a given condition, such as a specific ratio between the horizontal range and the maximum height. This article will walk you through the process of solving such a problem, where the horizontal range is four times the maximum height. We will use trigonometric identities and algebraic simplifications to find the angle of projection.

Understanding the Key Formulas

The horizontal range and maximum height of a projectile can be expressed as:

Formula for Maximum Height (H)

[ H frac{u^2 sin^2 theta}{2g} ] where ( u ) is the initial velocity, ( theta ) is the angle of projection, and ( g ) is the acceleration due to gravity.

Formula for Horizontal Range (R)

[ R frac{u^2 sin 2theta}{g} ] where ( u ) is the initial velocity, ( theta ) is the angle of projection, and ( g ) is the acceleration due to gravity.

Using Trigonometric Identities

We also have the identity for the double angle of sine: [ sin 2theta 2 sin theta cos theta ]

Setting Up the Problem

The problem states that the horizontal range is equal to four times the maximum height. We can set up the equation as follows: [ R 4H ] Substituting the formulas for ( R ) and ( H ), we get: [ frac{u^2 sin 2theta}{g} 4 left(frac{u^2 sin^2 theta}{2g}right) ] Simplifying, we get: [ frac{u^2 sin 2theta}{g} frac{4u^2 sin^2 theta}{2g} ] Dividing both sides by ( g ) and by ( u^2 ) (assuming ( u eq 0 )), we have: [ sin 2theta 2 sin^2 theta ] Using the identity ( sin 2theta 2 sin theta cos theta ), we rewrite the equation as: [ 2 sin theta cos theta 2 sin^2 theta ] Dividing both sides by 2 and assuming ( sin theta eq 0 ), we get: [ cos theta sin theta ] This simplifies to: [ tan theta 1 ] Therefore, the angle of projection is: [ theta 45^circ ]

Checking the Solution

To ensure our solution is correct, let's check the given conditions. For a projectile launched at an angle of ( 45^circ ):

Maximum Height (H)

[ H frac{u^2 sin^2 45^circ}{2g} frac{u^2 left(frac{1}{sqrt{2}}right)^2}{2g} frac{u^2 cdot frac{1}{2}}{2g} frac{u^2}{4g} ]

Horizontal Range (R)

[ R frac{u^2 sin 90^circ}{g} frac{u^2 cdot 1}{g} frac{u^2}{g} ] Thus, the horizontal range is indeed four times the maximum height: [ R 4H 4 left( frac{u^2}{4g} right) frac{u^2}{g} ] This confirms that the angle of projection is ( 45^circ ).

Alternative Solution: Using the Equation for Air Time

Another way to approach the problem is to consider the time of flight. The air time of a projectile is given by: [ t frac{2u sin theta}{g} ] The maximum height can also be expressed as: [ H frac{u^2 sin^2 theta}{2g} ] The horizontal range is then given by: [ R V cos theta cdot t V cos theta cdot frac{2u sin theta}{g} ] Given that ( R 4H ), we substitute the expressions for ( R ) and ( H ): [ frac{u^2 2 sin theta cos theta}{g} 4 left( frac{u^2 sin^2 theta}{2g} right) ] Simplifying, we get: [ 2 sin theta cos theta 2 sin^2 theta ] Dividing both sides by 2, we find: [ cos theta sin theta ] Thus, ( theta 45^circ ).

Conclusion

Summarizing, the angle of projection for a projectile whose horizontal range is four times its maximum height is ( 45^circ ). This solution not only applies the fundamental principles of projectile motion but also demonstrates the importance of trigonometric identities in solving such problems.

Key Takeaways:

The angle of projection is 45 degrees when the horizontal range is 4 times the maximum height. Use the formula for maximum height and horizontal range to set up the problem. Apply trigonometric identities and simplifications to solve for the angle.