Kvant Physics Problem 139

A ball is launched from a fixed point with speed $v$ at angle $\alpha = 30^\circ$ above the horizontal.

kvantphysics

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m09s
Source on kvant.digital

Problem

A footballer kicked a ball, giving it a speed $v$ at an angle $\alpha$ to the horizontal, and it went into the near lower corner of the goal. If the footballer had kicked the ball from the same point on the field and the ball had been launched 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 with which the ball initially starts moving, given that the height of the goal is $h=2~\text{м}$ and the angle is $\alpha=30^\circ$.

I. Sh. Slobodetsky

Setup and Assumptions

A ball is launched from a fixed point with speed $v$ at angle $\alpha = 30^\circ$ above the horizontal. Its trajectory passes through two different points of a goal depending on the initial speed.

For speed $v$, the ball reaches the lower corner of the goal, which lies on the ground level at height $y=0$ at the horizontal position of the goal. For speed $1.05v$, launched from the same point and at the same angle, the ball reaches the upper crossbar at height $y=h=2,\text{m}$ at the same horizontal position.

The motion is described in a fixed inertial frame attached to the ground. Air resistance is neglected. The gravitational acceleration is constant and equal to $g=9.81,\text{m/s}^2$. The ball is treated as a point mass.

The unknown is the initial speed $v$.

Physical Principles

The motion of a projectile under uniform gravity is governed by the kinematic equations

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

$$y(t)=v\sin\alpha , t - \frac{1}{2}gt^2.$$

Eliminating time gives the trajectory equation as a function of horizontal coordinate $x$:

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

At a fixed horizontal distance $x=L$, corresponding to the goal position, the height $y$ depends only on $v$.

Derivation

For the trajectory with speed $v$, the ball passes through the lower corner at $y=0$ when $x=L$. Substituting into the trajectory equation yields

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

Rearranging gives

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

Solving for $L$ yields

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

Using $\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$ gives

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

For the second trajectory with speed $1.05v$, the height at the same horizontal position $x=L$ equals $h$:

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

Substituting $L = \frac{v^2\sin 2\alpha}{g}$ into the first term gives

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

Using $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ leads to

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

For the second term, substituting $L^2 = \frac{v^4\sin^2 2\alpha}{g^2}$ gives

$$\frac{gL^2}{2(1.05v)^2\cos^2\alpha} = \frac{g}{2(1.05)^2 v^2 \cos^2\alpha}\cdot \frac{v^4\sin^2 2\alpha}{g^2}.$$

Simplifying yields

$$\frac{v^2\sin^2 2\alpha}{2g(1.05)^2\cos^2\alpha}.$$

Using $\sin^2 2\alpha = 4\sin^2\alpha\cos^2\alpha$ gives

$$\frac{2v^2\sin^2\alpha}{g(1.05)^2}.$$

Substituting both terms into the height condition gives

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

Factoring,

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

Solving for $v^2$ gives

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

Result

Substituting $h=2,\text{m}$, $g=9.81,\text{m/s}^2$, $\sin 30^\circ = \frac{1}{2}$, hence $\sin^2 30^\circ = \frac{1}{4}$:

$$v^2 = \frac{(2,\text{m})(9.81,\text{m/s}^2)}{2\cdot \frac{1}{4}\left(1 - \frac{1}{1.1025}\right)}.$$

Compute the numerical factor:

$$1.1025 = (1.05)^2, \quad \frac{1}{1.1025} \approx 0.9070,$$

$$1 - 0.9070 = 0.0930.$$

Thus

$$v^2 = \frac{19.62,\text{m}^2/\text{s}^2}{0.5 \cdot 0.0930} = \frac{19.62}{0.0465},\text{m}^2/\text{s}^2 \approx 422,\text{m}^2/\text{s}^2.$$

Taking the square root,

$$v \approx 20.5,\text{m/s}.$$

$$\boxed{v \approx 20.5,\text{m/s}}$$

Sanity Checks

The expression for $v^2$ contains the factor $\sin^2\alpha$ in the denominator, so decreasing the launch angle increases the required speed for the same target height difference, consistent with flatter trajectories dropping more strongly under gravity.

Dimensional analysis shows that $hg$ has units $\text{m}^2/\text{s}^2$, while the remaining factors are dimensionless, so $v^2$ has correct units of $\text{m}^2/\text{s}^2$.

The sensitivity to the factor $1-(1/1.05^2)$ is physically consistent, since for small percentage changes in speed the vertical deviation between trajectories is small; the denominator correctly becomes small, leading to a large inferred speed consistent with a goal height of order meters.