# Two projectiles at an initial height with a missing angle of projection

David and Goliath are testing what they learned in school, that the distance travelled by launching a projectile is dependent on a range of variables; the launch angle, height and velocity of projection.

As Goliath is significantly taller, David decides to throw a small projectile while standing on a ladder so that he is throwing from the same vertical height and same spot as Goliath.

Goliath launches first with an angle of 60$$^{\circ}$$ to the horizontal. David will then launch the projectile next with the same velocity of projection as Goliath.

If David wants to match Goliath’s attempt and achieve the same horizontal distance, at what angle should he launch the projectile if he can’t do it at an angle of 60$$^{\circ}$$?

(Use $$𝑔 = −10ms^{-1}$$ in your calculations and use this as the only force acting upon the projectile.)

If David's projectile's displacement is given by $$r_d = v t \cos \theta \hat{i} + (h + v t \sin \theta - 5t^2) \hat{j}$$ then David's projectile hits the ground when $$\frac{v \sin \theta \pm \sqrt{v^2 \sin^2\theta+20h}}{10} = 0$$. Solving this, the time it takes David's projectile to reach the ground is $$T_d = \frac{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}{10}$$.

If Goliath's projectile's displacement is given by $$r = \tfrac{v}{2} t \hat{i} + (h + \tfrac{\sqrt{3}v}{2}t - 5t^2) \hat{j}$$ then Goliath's projectile hits the ground when $$\frac{\sqrt{3}v \pm \sqrt{3v^2 + 80h}}{20} = 0$$. Solving this, the time it takes Goliath's projectile to reach the ground is $$T_g = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}$$.

For the horizontal distance travelled to match, their terminal horizontal components must match. Hence, \begin{align*} v T_d \cos \theta &= \tfrac{v}{2} T_g\\ T_d \cos \theta &= \tfrac{1}{2} T_g\\ \cos \theta &= \frac{T_g}{2} \frac{1}{T_d}\\ &= \frac{\frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}}{2} \frac{10}{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}\\ \cos \theta &= \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) }\\ \end{align*}

I'm having a major mental block on how to obtain an expression for $$\theta$$ in terms of $$h$$ and $$v$$. For what its worth, wolframalpha gives $$\theta = -\cos^{-1}\left(-\frac{\sqrt{200 h^2 + 70 h v^2 + 10 \sqrt{3} h v \sqrt{80 h + 3 v^2} + \sqrt{3} v^3 \sqrt{80 h + 3 v^2} + 3 v^4}}{\sqrt{800 h^2 + 80 h v^2 + 8 v^4}}\right)$$. I seem to missing out some details in my calculations.

What's the simplest way to obtain the formula for $$\theta$$ in terms of $$v$$ and $$h$$?

From $$\begin{equation*} \cos \theta = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) }, \end{equation*}$$ multiplying both numerator and denominator by $$v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h}$$ gives \begin{align*} \cos \theta &= \frac{(\sqrt{3}v + \sqrt{3v^2 + 80h})(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h})}{4 (-20h) }\\ &= A(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta}+20h}), \end{align*} where we have set $$A= -\dfrac{\sqrt{3}v + \sqrt{3v^{2} + 80h}}{80h}.$$ We then get \begin{align*} \Big(\frac{\cos{\theta}}{A} - v\sin{\theta}\Big)^{2} &= v^{2}\sin^{2}{\theta} + 20h, \end{align*} or equivalently \begin{align*} \frac{\cos^{2}{\theta}}{A^{2}} - 2\frac{v\cos{\theta}\sin{\theta}}{A} = 20h. \end{align*} Thus $$\begin{equation*} (\cos^{2}{\theta} - 20A^{2}h)^{2} = 4A^{2}v^{2}\cos^{2}{\theta}(1-\cos^{2}{\theta}). \end{equation*}$$ Expanding this equation gives the quadratic equation in $$\cos^{2}{\theta}$$ $$\begin{equation*} (1+4A^{2}v^{2})\cos^{4}{\theta}- 4A^{2}(10h + v^{2})\cos^{2}{\theta} + 400 A^{4}h^{2} = 0, \end{equation*}$$ and solving for $$\cos^{2}{\theta}$$ yields \begin{align*} \cos^{2}{\theta} &= \frac{4A^{2}(10h+v^{2})\pm\sqrt{16A^{4}(10h+v^{2})^{2}-4(400A^{4}h^{2})(1+4A^{2}v^{2})}}{2(1+4A^{2}v^{2})} \\[1ex] &= \frac{2A^{2}(10h+v^{2})\pm 2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}. \end{align*} There will be two solutions for $$\theta$$, one of them being at $$\theta = 60^{\circ}$$ which we aren't interested in, and the other being at some angle less than $$60^{\circ}$$. To get the smaller angle, we take the positive sign in the above expression since $$\cos^{2}{\theta}$$ is decreasing on $$\Big(0, \dfrac{\pi}{2}\Big)$$. Finally, as $$\cos{\theta} > 0$$, we obtain \begin{align*} \cos{\theta} = \sqrt{\frac{2A^{2}(10h+v^{2})+2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}}. \end{align*}

• I might have errors in redoing the calculations, but shouldn't $10h+v^2$ inside the square root be squared? – AB Balbuena Mar 9 at 10:50
• Also, is the term inside the square root always positive? I'm getting an expression for $\theta$ that may return an imaginary number – AB Balbuena Mar 9 at 20:35
• Yes it should be squared, thank you – I have updated my answer accordingly. This will allow for some simplification of what's inside the square root. As $A$ is a function of $v$ and $h$, the term may be negative for some range of values of $v$ and $h$. – Zac Mar 9 at 23:36

I got a very different answer.

https://en.wikipedia.org/wiki/Range_of_a_projectile

$$d = \frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right)$$

Now we must have

$$\frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right) = \frac{v^2 \sin 120^\circ}{g} = \frac{\sqrt{3}v^2}{2g}$$

Solving,

$$\theta = \frac{1}{2} \sin^{-1} \frac{3v}{\sqrt{6v^2 + 20y_0}}$$

• However, we're not interested in the range here. Only the measure of the angle that will have the same terminal point as throwing at 60degrees – AB Balbuena Mar 10 at 14:37
• Sorry, the question was "If David wants to match Goliath’s attempt and achieve the same horizontal distance, at what angle should he launch the projectile if he can’t do it at an angle of 60∘?" – PTDS Mar 10 at 17:32
• What is the difference between "horizontal distance" and "range"? – PTDS Mar 10 at 17:33

Let $$\theta$$ and $$v$$ be the launch angle and velocity, respectively. Then, the horizontal and vertical distances traveled are, $$x = v\cos\theta t,\>\>\>\>\> 0=v\sin\theta t - \frac 12 g t^2$$

Eliminate $$t$$ to get,

$$x= \frac{v^2}{g}\sin2\theta = \frac{v^2}{g}\sin(120^\circ)$$

where the given launch angle $$60^\circ$$ is used in the last step. Solve for $$\theta$$ of above equation to obtain the alternative launch angle at $$\theta = 30^\circ$$. (Note that the individuals' height is assumed to be small compared to the distance that the projectile travels and is negligible.)

• The whole question is phrased in a way that the initial height of projection matters though. "...the distance travelled by launching a projectile is dependent on a range of variables; the launch angle, height and velocity of projection. " – AB Balbuena Mar 6 at 23:16
• @ABBalbuena - I made the assumption based on the problem specification of a small projectile. Good luck. – Quanto Mar 10 at 2:32