How to find the equation of a parabola when only given the x-intercepts and the Axis of symmetry?

Question

i have been given the following problem.

A tennis ball is lobbed from ground level and must cover a horizontal distance of 22m if it is to land just inside the opposite end of the court. If the opponent is standing 4m from the baseline and he can hit any ball less than 3m high, what is the lowest maximum height the lob
must reach to win the point.

The answer is 5 and 1/24 Metres (back of the textbook) I have no idea how to approach this problem but my attempt to
approach this problem is by finding the X-intercepts which are 0, 22. Thus, finding the Axis of symmetry that is 22/2 = 11. That is all i've done so far.

Explanation will be extremely helpful :)

Answer 1

First in order to have an equation you must have a coordinate system! I suspect that you are taking the position from which the ball is lobbed to be then "0" point, you want the parabola to pass through (0, 0) and (22, 0). You want the y value, when x= 22- 4= 18, to be at least 3. Writing the parabola as $y= ax^2+ bx+ c$ we must have (a) $0= a(0^2)+ b(0)+ c= c$. (b)$$0= a(22^2)+ b(22)$ and (c) $y= a(18^2)+ b(18)\ge 3$. From the second equation, b= -22a. The last inequality becomes then, $324a- 432\ge 3$ or $324a\ge 435$. So $a\ge 435/324= 4/3$ and b= -22a\ge 88/3.

From $y= ax^2+ bx$, the highest point comes where $y'= 2ax+ b= 2ax- 22a= 0$ so $a= 22/2= 11$ as you say. And that highest value is $y= 121a- 22a(11)= 121a- 242a= -121a (of course, since the parabola arches upward a< 0).

Answer 2

The question can be solved mentally with a bit of creativity.

Setting $x=0, y = h$ for the vertex, the ball travels $7$ ft horizontally to drop to a height of $3$ ft,
and a further $4$ ft horizontally to drop to the ground.

Since the horizontal velocity of a projectile (w/o resistance) is constant,
the ratios of the distances exactly depict the ratio of the times taken to drop.

And since $h \propto t^2,\;\; \dfrac{h}{3} = \left(\dfrac{11^2}{11^2-7^2}\right) \to h = 5\frac1{24}$

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Added

It has been taken as if a ball has been dropped from rest from a height $h$

So if we take it that a time of $11$ is taken for the ball to reach the ground, a time of $7$ will be needed to reach $3$ ft above the ground.

The distance fallen $s \propto t^2$, and we are only using ratios, so we can take the corresponding distances to be $11^2$ and $7^2$.

So $11^2$ represents the full height $h$, $11^2-7^2$ represents $3$ ft, and we directly get

$\dfrac{h}{3} = \left(\dfrac{11^2}{11^2-7^2}\right) \to h = 5\frac1{24}$

Answer 3

The general equation of a parabola is $$ f(x) = a x^2 + bx + c $$ it has three parameters.

We know $f(11) = 0$, so we have $$ f(11) = 0 $$ we also know $$ 3 = f(11-4) = f(7) $$ further we lobbed from ground level, such that we are 22m away, so this means $$ 0 = f(-11) = a 11^2 - 11 b + c $$ This gives the linear system in the unknowns $a,b,c$: $$ \left[ \begin{array}{rrr|r} 121 & 11 & 1 & 0 \\ 49 & 7 & 1 & 3 \\ 121 & -11 & 1 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|r} 242 & 0 & 2 & 0 \\ 49 & 7 & 1 & 3 \\ 121 & -11 & 1 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 1/121 & 0 \\ 49 & 7 & 1 & 3 \\ 0 & -11 & 0 & 0 \end{array} \right] \to \\ \left[ \begin{array}{rrr|r} 1 & 0 & 1/121 & 0 \\ 0 & 1 & 0 & 0 \\ 49 & 0 & 1 & 3 \\ \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 1/121 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1-49/121 & 3 \\ \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 1/121 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 72/121 & 3 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrr|r} 1 & 0 & 0 & -1/24 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 121/24 \\ \end{array} \right] $$

The highest point is achieved in the middle, having height $$ f(0) = c = 121/24 $$

Answer 4

Let $H$ be the maximum height achieved.

Imagine that you are viewing the tennis game upside down.

Let the origin $O (0,0)$ be the highest (lowest when upside down) point of the ball. The trajectory of the ball is then a parabola opening upwards with equation $y=ax^2$ and passes through $(7,H-3)$ and $(11,H)$, i.e. $$\begin{align}\begin{cases}H-3&=a(7)^2\\ H&=a(11)^2\end{cases}\\ 1-\frac 3H&=\frac {49}{121}\\ H&=\frac {121}{24}=5\frac 1{24}\qquad\blacksquare\end{align}$$