Food for thought

## Solution

A ball is thrown from a position $O$ on level ground with initial velocity (in $\mathrm{m\,s^{-1}}$) $\mathbf{u}=5\mathbf{i}+6\mathbf{j}$ where $\mathbf{i}$ is a horizontal direction and $\mathbf{j}$ is vertically upwards. …

One way to model the effect of the wind is as a constant horizontal acceleration. Let’s call the magnitude of this acceleration $\quantity{w}{m\,s^{-2}}$.

Find how far from $O$ the ball will land as a function of $w$. Express the trajectory of the ball using parametric equations for $x$ and $y$ in terms of time, $t$.

Take gravity to be $\quantity{10}{m\,s^{-2}}$.

The horizontal and vertical motions are independent so we can treat them separately. Vertically, the wind has no effect so the ball has constant acceleration under gravity. $$$y = 6t-\frac{1}{2}10t^2 = 6t-5t^2 \label{eq:vert}$$$ Horizontally, the ball has initial velocity $5$ and constant acceleration $-w$. So we have $$$x = 5t-\frac{1}{2}w t^2 = 5t-\frac{w}{2}t^2. \label{eq:horiz}$$$ To find where the ball lands, we set $y=0$ which means \begin{align*} 6t-5t^2 &= 0 \\ \implies\quad t&= 0 \text{ or } \frac{6}{5}. \end{align*}

Note that this is exactly the same as we had in the Warm-up. Why is that?

Substituting $t=\frac{6}{5}$ into equation $\eqref{eq:horiz}$ gives us the landing distance as a function of $w$: $$$x_l=6-\frac{18}{25}w. \label{eq:x-land}$$$

The equations $\eqref{eq:vert}$ and $\eqref{eq:horiz}$ are a parametric form of the trajectory.

In each of these cases, find where the ball lands and draw a sketch of the trajectory:

• $w=0$

This is the case where the wind has no effect so the behaviour is exactly as in the Warm-up. The trajectory is a parabola with maximum at $\left(3,\frac{9}{5}\right)$. The ball lands at $x_l=6$.

In each of these cases, find where the ball lands and draw a sketch of the trajectory:

• $w=2.5$

Substituting $w=\frac{5}{2}$ into $\eqref{eq:x-land}$ we find that the ball lands at $x_l=6-\frac{18}{25}\times\frac{5}{2} = 4.2,$ so the wind has reduced the range from $\quantity{6}{m}$ to $\quantity{4.2}{m}$.

The trajectory is given by equations $\eqref{eq:vert}$ and $\eqref{eq:horiz}$ as \begin{align*} x &= 5t-\frac{5}{4}t^2 \\ y &= 6t-5t^2. \end{align*}

To help sketch the curve it is useful to locate the highest point on the trajectory. The height is a quadratic function of $t$ and the ball lands when $t=\frac{6}{5}$ so it is at it’s maximum height when $t=\frac{3}{5}$. Substituting in the above gives us the highest point at $x=\frac{51}{20} = 2.55, \quad y=\frac{9}{5}=1.8.$

Notice that the maximum occurs at the same height and time as in the Warm-up, but that is has moved horizontally back towards the thrower. Also, the ball travels less far horizontally in the second half of its trajectory than it did in the first half.

In each of these cases, find where the ball lands and draw a sketch of the trajectory:

• $w=5$

The landing point is at $x_l=6-\frac{18}{5} = 2.4.$

The trajectory is \begin{align*} x &= 5t-\frac{5}{2}t^2 \\ y &= 6t-5t^2. \end{align*}

Both $x$ and $y$ are quadratic functions of $t$ and they each have a maximum value. The maximum $y$ always occurs when $t=\frac{3}{5}$. We can find when the maximum $x$ occurs by completing the square or by comparing the roots – in this case it is when $t=1$. So the maximum $y$ is at $(2.1,1.8)$ and maximum $x$ is at $(2.5,1)$.

The ball starts to move back towards the thrower before it lands.

In each of these cases, find where the ball lands and draw a sketch of the trajectory:

• $w=10$

The landing point is at $x_l=6-\frac{18}{25}\times 10 = -1.2.$

The trajectory is \begin{align*} x &= 5t-5t^2 \\ y &= 6t-5t^2 \end{align*}

which has maximum $y$ at $(1.2,1.8)$ and maximum $x$ at $(1.25,1.75)$.

The ball has been blown back past the thrower.

What stays the same and what changes as we vary the value of $w$?

What value of $w$ would make the ball return to $O$?

This applet shows how the trajectory varies.

• The shape of the trajectory is always a parabola, but it’s width and orientation change.
• The ball always reaches the same maximum height.
• The landing point gets closer to $O$ as $w$ increases until the point where it goes behind the thrower.

The ball returns to $O$ if $x_l=0$ which means $6-\frac{18}{25}w = 0 \quad\implies\quad w=\frac{25}{3}.$

What shape is the trajectory in this case?

What are the limitations of this model?

The model of constant $w$ is quite a bold simplification of reality. The acceleration is actually some function of the ball’s velocity relative to the surrounding air. We could say $w=f\left(v_\text{wind} - v_\text{ball}\right).$ In the general case these would all be vector quantities, but here we are only considering the horizontal components. If the wind speed is constant and significantly bigger than the speed of the ball then the input to the function $f$ does not change much as the ball speeds up or slows down. Hence we can say that $w$ will not change very much, which is how we could justify a constant $w$ model.

In practice, the effect of the wind will almost certainly not be a constant acceleration. A more accurate model of this, however, would lead to some equations that are hard to solve.

In addition to the horizontal component of acceleration caused by the wind, there would be a vertical component which we might think of as air resistance. Again, this is hard to model accurately.

Other problems with this model include

• the wind doesn’t usually blow at a constant speed,
• the wind speed often varies with height above the ground,
• any spin on the ball could alter its trajectory.