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Q42E

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Found in: Page 125

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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 40º above the horizontal.

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

See the step by step solution

## Step 1: Given data

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

## Step 2: Initial velocity of the ball

Figure: The motion of the ball

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

$\begin{array}{rcl}{c}{o}{s}{\theta }& {=}& \frac{a}{h}\\ \mathrm{cos}40& =& \frac{{V}_{ix}}{{V}_{i}}\\ {V}_{ix}& =& {V}_{i}\mathrm{cos}{40}^{°}\end{array}$

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

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

$\begin{array}{rcl}{s}{i}{n}{\theta }& {=}& \frac{o}{h}\\ \mathrm{sin}40& =& \frac{{V}_{iy}}{{V}_{i}}\\ {V}_{iy}& =& {V}_{i}\mathrm{sin}40\end{array}$

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:

$\begin{array}{rcl}\overline{)V}& =& \frac{x}{t}\\ {V}_{ix}& =& \frac{30}{t}\\ t& =& \frac{30}{{V}_{i}\mathrm{cos}40}\end{array}$

The time taken is used from the above equation.

From the equation of motion:

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

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

$\begin{array}{rcl}2.4& =& \left({V}_{i}\mathrm{sin}40\right)\left(\frac{30}{{V}_{i}\mathrm{cos}40}\right)+\frac{1}{2}\left(-9.8\right){\left(\frac{30}{{V}_{i}\mathrm{cos}40}\right)}^{2}\\ 2.4& =& 25.2-\frac{4410}{{{V}_{i}}^{2}×0.587}\\ \frac{4410}{{{V}_{i}}^{2}×0.587}& =& 25.2-2.4\\ {{V}_{i}}^{2}& =& \frac{7510}{22.8}\\ {{V}_{i}}^{2}& =& 329\\ {V}_{i}& =& 18.1m}{s}\\ & & \end{array}$

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