Mv6, Mv7 - Projectiles on an inclined plane

Mv6, Mv7 - Projectiles on an inclined plane

Find where a projectile next meets a uniform slope and choose the launch direction that maximises its range. You will distinguish angles from the horizontal and from the slope, resolve gravity into rotated coordinates, and measure range along the plane.

Coordinates and angles on a slope

Is 45∘45^\circ above horizontal still the best launch angle when the landing surface rises at 30∘30^\circ?

The landing surface changes the answer. A steeper upward launch can keep the projectile above that rising surface for longer. We will derive the best angle rather than guess it.

You already know

A projectile is modelled as a particle with no air resistance and constant gravitational acceleration g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} vertically downwards. The distances must be small enough for gravity's direction and magnitude to remain constant.

Let the slope rise at signed angle α\alpha as you move right: α>0\alpha>0 is uphill and α<0\alpha<0 downhill, with −90∘<α<90∘-90^\circ<\alpha<90^\circ. Take xx along that direction of the slope and yy perpendicular above it. Launch is from the slope itself. Let β\beta be the launch angle above the slope, so the horizontal launch angle is θ=α+β\theta=\alpha+\beta.

[DIAGRAM: asset_name: Projectile coordinates on an inclined plane; asset_slug: inclined_projectile_axes; caption: The launch angle above the horizontal is alpha plus beta; description: A plane rising 30 degrees has a blue x-axis along it and sage y-axis normal above it. A launch arrow at 60 degrees above horizontal makes beta 30 degrees with the plane. A downward gravity arrow is resolved into negative along-slope and negative normal components. The two angle arcs have separate labels.]
The launch angle above the horizontal is alpha plus beta

Coordinates parallel and normal to a slope

For launch speed u>0u>0,

ax=−gsin⁡α,ay=−gcos⁡αx=ucos⁡β t−12gsin⁡α t2y=usin⁡β t−12gcos⁡α t2\begin{aligned} a_x&=-g\sin\alpha,&a_y&=-g\cos\alpha\\ x&=u\cos\beta\,t-\frac12g\sin\alpha\,t^2\\ y&=u\sin\beta\,t-\frac12g\cos\alpha\,t^2 \end{aligned}

Use 0<β<90∘0<\beta<90^\circ for an initially forward launch above the plane, and require a forward landing in the chosen direction.

These resolved equations are derived from the constant-acceleration formulae; they are not printed separately.

Not in the formula booklet: learn it.

When α<0\alpha<0, the along-slope acceleration is positive: gravity helps motion downhill. The normal acceleration remains towards the plane.

Key idea

The slope angle α\alpha, launch angle above the slope β\beta, and launch angle above horizontal θ=α+β\theta=\alpha+\beta are different quantities.

Resolve gravity before calculating a range.

Finding the next landing point

The next landing has perpendicular coordinate y=0y=0; its along-slope coordinate is the range RR.

Factoring the normal-coordinate equation gives the launch root t=0t=0 and the positive flight time:

t=2usin⁡βgcos⁡αt=\frac{2u\sin\beta}{g\cos\alpha}

Substitution into the along-slope equation, followed by
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta, gives the range.

Range along a uniform slope

With the launch point on the slope,

R=2u2sin⁡βcos⁡(α+β)gcos⁡2αR=\frac{2u^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}

This is range along the plane; its horizontal component is Rcos⁡αR\cos\alpha. Under the stated coordinate convention, require R>0R>0, equivalently cos⁡(α+β)>0\cos(\alpha+\beta)>0.

Not in the formula booklet: learn it.

You can instead use horizontal and vertical coordinates and set vertical height equal to the slope's height. Either approach must measure the final range along the plane.

Worked example 1: an uphill landing

Question. A particle is projected at 14 m s−114\,\mathrm{m\,s^{-1}}, 30∘30^\circ above a plane rising at 30∘30^\circ. Find its range up the plane.

Plan. Use α=β=30∘\alpha=\beta=30^\circ, discard the zero-time root, and calculate along-slope range.

t=2(14)(1/2)9.8(3/2)=2073positive flight timeR=2(142)(1/2)cos⁡60∘9.8(3/4)slope-range formula=403simplify\begin{aligned} t&=\frac{2(14)(1/2)}{9.8(\sqrt3/2)} =\frac{20}{7\sqrt3} && \text{positive flight time}\\ R&=\frac{2(14^2)(1/2)\cos60^\circ}{9.8(3/4)} && \text{slope-range formula}\\ &=\frac{40}{3} && \text{simplify} \end{aligned}

Why this step? The cosine uses the total horizontal launch angle 60∘60^\circ, not the angle 30∘30^\circ above the plane.

Answer. 40/3 m40/3\,\mathrm{m}, about 13.3 m13.3\,\mathrm{m}, along the slope.

Check. Direct substitution gives x=73 t−2.45t2=20−20/3=40/3x=7\sqrt3\,t-2.45t^2=20-20/3=40/3.

Reverse only the slope angle to −30∘-30^\circ, retaining β=30∘\beta=30^\circ. The particle is now launched horizontally above a descending plane.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account