I got stuck on exercise 26, 566p in calculus by morris kline.
- A projectile is fired at an initial velocity of 160 ft/sec. Find the angle of fire $A$ so that the projectile will strike a wall 480 feet away at the maximum possible height.
It's solution to the problem.
$y = -\frac{16}{V^2\cos^2A}x^2 + x\tan A.\tag{10}$
From (10) with $V = 160$ and $x = 480$, we find the height $y$ is given in terms of the firing angle $A$ by $y = -144\sec^2A + 480\tan A$. Setting $\frac{dy}{dA} = 0$ and noting that $\sec^2A \not= 0$, we obtain the condition $\tan A = \frac{5}{3}$. Thus $A = 59\unicode{xB0}$(approx.).
The logic of the solution is followable. But when a projectile is at the max height, $x$ coordinate is one half of the range. Range of certain projectile with initial velocity $V$ and inclination $A$ is $\frac{V^2}{32}\sin 2A$. This implies with $V = 160$, range is at most 800 and therefore $x$ coordinate with maximum height is at most 400. But the exercise gives 480 which is greater than 400. So inserting $V = 160$ and $x = 480$ in (10) is nonsense. Is the exercise wrongly designed? Or am I misunderstanding?
Best Answer
I believe the solution they have provided is correct!
They have clearly mentioned that there is a wall at $x = 480$ So, now you should make this clear that the maximum height of the projectile should reach at $x = 480$.
Now, They want you to have a projectile angle A for which your projectile will have max. height at $x=480$. that does not mean that the height of projectile @$x=480$ should exceed the height of projectile @$x=400$
The above graph shows traces of points (Ay < By)
Of course!, As you said the ymx at x=400 > ymx at x=480 is right.
But they only mean the maximum possible height at which the projectile can hit the wall.
Means,
say at angle $A = $a1 ymx at x=480 $=$y1
keep doing this way now for what value of angle $A$ you'll hit the wall max
Which Of course! the value of ymx (x,A)=(480,a) will never exceed ymx @x=400 for whatever value of angle $A$