Question

1. can a goalkeeper at her/ his goal kick a soccer ball into the opponents goal without the ball touching the ground? the distance will be about 95 m. a goalkeeper can give the ball a speed of 30 m/s.
2. derive the range of a projectile on level ground in terms of initial velocity and launch angle.

Explanation:

Step1: Recall range formula for projectile

The range formula for a projectile on level - ground is $R=\frac{v_0^{2}\sin2\theta}{g}$, where $v_0$ is the initial velocity, $\theta$ is the launch angle, and $g = 9.8\ m/s^{2}$ is the acceleration due to gravity.

Step2: Analyze the maximum range

The function $y = \sin2\theta$ has a maximum value of 1 when $2\theta=90^{\circ}$ or $\theta = 45^{\circ}$.

Step3: Calculate the maximum range for the given initial velocity

Given $v_0 = 30\ m/s$, substituting into the range formula with $\sin2\theta = 1$ (at $\theta = 45^{\circ}$), we get $R=\frac{v_0^{2}}{g}$.
$R=\frac{30^{2}}{9.8}=\frac{900}{9.8}\approx91.84\ m$.

Step4: Answer the first question

Since the maximum range with an initial speed of $30\ m/s$ is approximately $91.84\ m$ and the distance to the opponent's goal is about $95\ m$, a goalkeeper cannot kick a soccer - ball into the opponent's goal without the ball touching the ground.

Step5: Derive the range formula for the second question

1. Break the initial velocity into components:

The initial velocity $v_0$ can be broken into horizontal $v_{0x}=v_0\cos\theta$ and vertical $v_{0y}=v_0\sin\theta$ components.

1. Find the time of flight:

The vertical displacement $y - y_0 = 0$ (level - ground). Using the equation $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, we have $0 = v_0\sin\theta t-\frac{1}{2}gt^{2}$. Factoring out $t$, we get $t(v_0\sin\theta-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the initial time). The other solution for the time of flight $T$ is $T=\frac{2v_0\sin\theta}{g}$.

1. Find the range:

The horizontal displacement $R$ (range) is given by $R = v_{0x}T$. Substituting $v_{0x}=v_0\cos\theta$ and $T=\frac{2v_0\sin\theta}{g}$, we get $R=\frac{v_0\cos\theta\times2v_0\sin\theta}{g}$. Using the double - angle formula $\sin2\theta = 2\sin\theta\cos\theta$, we can rewrite the range formula as $R=\frac{v_0^{2}\sin2\theta}{g}$.

Answer:

1. No.
2. $R=\frac{v_0^{2}\sin2\theta}{g}$