Kvant Physics Problem 139

Verified: yes
Verdicts: PASS + PASS
Solve time: 3m08s
Source on kvant.digital

Problem

A footballer kicked the ball, imparting to it a speed $v$ at an angle $\alpha$ to the horizontal, and it reached the near lower corner of the goal. If the footballer had kicked the ball from the same point on the football field and the ball had flown at the same angle to the horizontal, but with a speed 5% greater than $v$, it would have hit the upper crossbar of the goal. Find the speed at which the ball starts moving, if the height of the goal is $h=2~\text{м}$ and the angle is $\alpha=30^\circ$.

I. Sh. Slobodetskii

Setup and Assumptions

A football is launched from a fixed point on level ground with initial speed $v$ at an angle $\alpha=30^\circ$ to the horizontal. The ball first lands at the near lower corner of the goal, which is taken to lie on the ground level, and this horizontal position is denoted by $x=L$. The height of the goal crossbar is $h=2,\text{m}$.

In a second scenario the launch point and angle are unchanged, while the speed becomes $1.05v$. The horizontal position of the goal is the same, so the projectile is evaluated at the same coordinate $x=L$, and in this case the ball reaches height $y=h$.

Air resistance is neglected, the gravitational acceleration is constant and equal to $g$, and the motion is treated in a fixed inertial reference frame attached to the ground. The ball is modeled as a point particle.

The unknown quantity is the initial speed $v$.

Physical Principles

The motion of a projectile under uniform gravity is described by the parametric equations

$x(t)=v\cos\alpha , t,$

$y(t)=v\sin\alpha , t-\frac{g t^2}{2}.$

Eliminating time gives the trajectory equation

$y(x)=x\tan\alpha-\frac{g x^2}{2 v^2 \cos^2\alpha}.$

At a fixed horizontal position $x=L$, the vertical coordinate depends on the launch speed $v$ according to this relation.

Derivation

For the first kick, the ball reaches the ground at the lower corner of the goal, so at $x=L$ the vertical coordinate is zero. Substituting $x=L$ and $y=0$ into the trajectory equation gives

$0=L\tan\alpha-\frac{g L^2}{2 v^2 \cos^2\alpha}.$

Rearranging yields

$L\tan\alpha=\frac{g L^2}{2 v^2 \cos^2\alpha}.$

Dividing by $L$ gives

$\tan\alpha=\frac{g L}{2 v^2 \cos^2\alpha},$

and solving for $L$ produces

$L=\frac{2 v^2 \cos^2\alpha \tan\alpha}{g}=\frac{2 v^2 \sin\alpha \cos\alpha}{g}=\frac{v^2 \sin 2\alpha}{g}.$

For the second kick, the speed is $1.05v$, and at the same horizontal distance $x=L$ the ball reaches the crossbar height $h$. Substitution into the trajectory equation gives

$h=L\tan\alpha-\frac{g L^2}{2 (1.05v)^2 \cos^2\alpha}.$

From the first motion, the identity

$L\tan\alpha=\frac{g L^2}{2 v^2 \cos^2\alpha}$

is available, so the expression for $h$ becomes

$h=\frac{g L^2}{2 v^2 \cos^2\alpha}\left(1-\frac{1}{(1.05)^2}\right).$

Substituting $L=\frac{v^2 \sin 2\alpha}{g}$ gives

$h=\frac{g}{2 v^2 \cos^2\alpha}\cdot \frac{v^4 \sin^2 2\alpha}{g^2}\left(1-\frac{1}{(1.05)^2}\right).$

Simplifying,

$h=\frac{v^2 \sin^2 2\alpha}{2 g \cos^2\alpha}\left(1-\frac{1}{(1.05)^2}\right).$

Using $\sin 2\alpha=2\sin\alpha\cos\alpha$, the factor simplifies as

$\frac{\sin^2 2\alpha}{\cos^2\alpha}=4\sin^2\alpha,$

so the expression becomes

$h=\frac{2 v^2 \sin^2\alpha}{g}\left(1-\frac{1}{(1.05)^2}\right).$

Solving for $v^2$ gives

$v^2=\frac{h g}{2 \sin^2\alpha \left(1-\frac{1}{(1.05)^2}\right)}.$

Result

Substituting $h=2,\text{m}$, $\alpha=30^\circ$ with $\sin 30^\circ=\frac{1}{2}$, and taking $g=10,\text{m/s}^2$, first evaluate

$\sin^2\alpha=\frac{1}{4},$

$1-\frac{1}{(1.05)^2}=1-\frac{1}{1.1025}\approx 0.09297.$

Then

$v^2=\frac{(2,\text{m})(10,\text{m/s}^2)}{2\cdot \frac{1}{4}\cdot 0.09297}=\frac{20,\text{m}^2/\text{s}^2}{0.5\cdot 0.09297}=\frac{20}{0.046485},\text{m}^2/\text{s}^2.$

This gives

$v^2\approx 4.30\times 10^2,\text{m}^2/\text{s}^2,$

so

$v\approx 20.7,\text{m/s}.$

$$\boxed{v \approx 2.1\times 10^{1}\ \text{m/s}}$$

Sanity Checks

The expression for $v^2$ has dimensions determined by $\frac{h g}{1}$, since the trigonometric and numerical factors are dimensionless, giving $\text{m}\cdot \text{m/s}^2=\text{m}^2/\text{s}^2$, consistent with $v^2$.

If the increment in speed were reduced so that $1.05v\to v$, the factor $1-\frac{1}{(1.05)^2}$ would tend to zero, forcing $v\to\infty$, which matches the requirement that two distinct heights at the same range cannot be produced by identical trajectories.

The dependence $v\propto \frac{1}{\sin\alpha}$ is consistent with the role of vertical velocity in reaching the height difference between ground level and the crossbar.