Problem with logic regarding projectile motion problems when only Range and Angle are given

Question

Im struggling with understanding a problem with my logic in regards to solving problems relating to finding out an initial velocity when given a range $d$ and angle of elevation $\theta$. The main method involves using the equations $$v_i \cos(\theta)t=d$$ and $$v_{fy}=v_i\sin\theta\ +gt$$ The method involves substitution with the time variable and solving for $v_i$. I however found a line of reasoning that gives the incorrect answer by a large amount. I am confused about what exactly is making it incorrect.

My line of reasoning is dividing the range by $2$ to form a right triangle with the maximum height and using $\frac{1}{2}d\tan\theta$ to find the maximum height $h$. I then use the equation $$v_f^2=v_i^2+2ad$$ to find the initial velocity for the $y$ component using $h$ for $d$, the acceleration of gravity for $a$, and $v_f=0$ as its the apex of the travel. then using $v_i = v_{iy}/\sin\theta$ I find the initial velocity. This answer is always very very wrong but I just don't see where the problem resides in this logic. if you could help explain this it would be very helpful as I can’t seem to find a good explanation from my teachers or online.

Answer 1

Your problem is that using a right triangle here is invalid. Because the projectile flies in a parabolic arc, you are finding a meaningless value for $h$ by assuming that the trajectory carries on in a straight line to the apex.

Answer 2

In general, when you have one method that works and are trying to understand the error of an alternative method, you can often use the working method to examine the assumptions you made in the other method, and that in turn can often help you to pinpoint the error.

Here, you make the assumption about the relation of the maximum height the projectile attains, $h_{\text{max}}$ to the projection angle $\theta$ and the range $d$. So you can check what is the correct relation by using the method that you know is correct.

For example, one way that you can find the maximum height is by energy conservation:

$$ \frac{1}{2}mv_i^2\sin^2\theta = mgh_{\text{max}}$$

$$ h_{\text{max}} = \frac{v_i^2\sin^2\theta}{2g}$$

Now the time to reach maximum height is given by:

$$ v_{fy} = 0 = v_i\sin\theta-gt_{\text{top}}$$ $$ t_{\text{top}}=\frac{v_i\sin\theta}{g}$$ By symmetry of the parabolic path, the entire distance is covered in twice that duration. So:

$$ d = v_i\cos\theta \cdot 2t_{\text{top}} = \frac{2v_i^2 \sin\theta\cos\theta}{g} \qquad (\ast) $$

Now if we calculate $\frac{1}{2}d\tan\theta$, as you suggested:

\begin{align*} \frac{1}{2}d\tan\theta &= \frac{1}{2}\frac{2v_i^2 \sin\theta\cos\theta}{g}\tan\theta \\ &= \require{cancel}\frac{v_i^2 \sin\theta\cancel{\cos\theta}}{g}\frac{\sin\theta}{\cancel{\cos\theta}} \\ &= \frac{v_i^2\sin^2\theta}{g} \end{align*}

Which turns out to be precisely $2h_\text{max}$. This isn't surprising, the height of a right triangle with angle $\theta$ and base $d/2$ is higher than the actual height attained by the projectile, because the projectile would follow a straight line only in the absence of gravity (and would never descend).

We see however that it would be correct to write:

$$ h_{\text{max}} = \frac{1}{4}d\tan\theta $$

So the actual maximum height attained by the projectile is half the height of a right triangle with base $d/2$ and base angle $\theta$. This is actually a well known result, as you can see for example here.

It's nice to note that there is no mention of $g$ in this relation. The effect of $g$ is embedded in the value of $d$; a stronger gravitational acceleration will reduce $d$ and vice versa. In fact you can see that when $g\rightarrow 0$, the expression given in $(\ast)$ gives $d\rightarrow\infty$. As the gravitational acceleration vanishes the maximum range extends without bound.