Question

Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40ºabove the horizontal.

Short answer

The initial velocity of the ball will be 18.1 m/s.

Step-by-step solution

Given data

• The change in the displacement in the y-direction is 2.4 meters.
• The acceleration in the above case is -9.8m/s².

Initial velocity of the ball

Figure: The motion of the ball

The initial velocity in the x-direction can be caculated as:

$\begin{array}{rcl}cos\theta & =& \frac{a}{h}\\ \cos 40& =& \frac{V_{ix}}{V_i}\\ V_{ix}& =& V_i\cos {40}^{°}\end{array}$

The initial velocity in the x-direction is the above equation.

The initial velocity in the y-direction can be caculated as:

$\begin{array}{rcl}sin\theta & =& \frac{o}{h}\\ \sin 40& =& \frac{V_{iy}}{V_i}\\ V_{iy}& =& V_i\sin 40\end{array}$

The initial velocity in the y-direction is above the equation.

The velocity in the x frame will be constant at all times. Considering the initial velocity, we can write:

$\begin{array}{rcl}\overline{V}& =& \frac{x}{t}\\ V_{ix}& =& \frac{30}{t}\\ t& =& \frac{30}{V_i\cos 40}\end{array}$

The time taken is used from the above equation.

From the equation of motion:

$\mathbf{∆}\mathit{Y}\mathbf{=}{\mathit{V}}_{\mathbf{i}\mathbf{y}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$

By putting the values in the above equation, we get:

$\begin{array}{rcl}2.4& =& \left(V_i\sin 40\right)\left(\frac{30}{V_i\cos 40}\right)+\frac{1}{2}\left(-9.8\right){\left(\frac{30}{V_i\cos 40}\right)}^2\\ 2.4& =& 25.2-\frac{4410}{{V_i}^2\times 0.587}\\ \frac{4410}{{V_i}^2\times 0.587}& =& 25.2-2.4\\ {V_i}^2& =& \frac{7510}{22.8}\\ {V_i}^2& =& 329\\ V_i& =& 18.1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ & & \end{array}$

Therefore, the initial velocity will be 18.1 m/s.