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Q47PE

Expert-verifiedFound in: Page 124

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A football player punts the ball at a $\mathbf{45}.\mathbf{0}$angle. Without an effect from the wind, the ball would travel $\mathbf{60}.\mathbf{0}\text{}\mathbf{m}$horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by $\mathbf{1}.\mathbf{50}\text{}\mathbf{m}/\mathbf{s}$. What distance does the ball travel horizontally?**

(a) Initial speed of the ball is $24.2m/s.$

(b) The total displacement will be $57.2$meter.

**When gravity first exerts force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.**

The range of the football is 60 m.

The angle of the football is released at the angle 45 degree.

The gravitational acceleration is -9.8 m/s^{2}

The initial velocity we have to calculate.

The velocity in the x frame will be constant in all time. Considering the initial velocity

$\begin{array}{rcl}{R}_{x}& =& \frac{{{V}_{i}}^{2}sin2{\theta}_{i}}{g}\\ 60& =& \frac{{{V}_{i}}^{2}sin2\left(45\right)}{9.8}\\ {V}_{i}& =& \sqrt{588}\\ {V}_{i}& =& 24.2\u200am/s\\ & & \end{array}$

The initial velocity will be positive, that is 24.2 m/s.

The Components of the initial velocity.

$\begin{array}{rcl}{V}_{ix}& =& {V}_{i}cos\theta \\ {V}_{ix}& =& \left(24.2\right)cos45\\ {V}_{ix}& =& 17.1\u200am/s\\ & & \end{array}$

The initial velocity in the x frame for part a is $17.1\u200am/s$.

The velocity in the direction for frame B will be

$\begin{array}{rcl}{V}_{xb}& =& 17.1-1.50\\ & =& 15.6\u200am/s\\ & & \end{array}$

Time is needed for finding the distanceThe Y- Components of the initial velocity.

$\begin{array}{rcl}{V}_{iy}& =& {V}_{i}sin\theta \\ {V}_{iy}& =& \left(24.2\right)sin45\\ {V}_{iy}& =& 17.1\u200am/s\\ & & \end{array}$

The initial velocity in the y frame for part a is -$17.1\u200am/s$. As the ball is decreasing so value of velocity became negative.

Putting the value of the given data in the equation

$\begin{array}{rcl}{V}_{f}& =& {V}_{i}+at\\ -17.1& =& 17.1+\left(-9.8\right)t\\ t& =& 3.49\u200as\\ & & \end{array}$

The time is total time 3.49 s, but for going from O to A and then from A to B the time becomes half ${t}_{a}=\frac{3.49}{2}=1.75\u200a\u200as$

The displacement for part A is

$\begin{array}{rcl}\overline{{V}_{a}}& =& \frac{{X}_{a}}{{t}_{a}}\\ {X}_{a}& =& 17.1\times 1.75\\ {X}_{a}& =& 29.9\u200am\\ & & \end{array}$

The displacement for part A is $29.9$meter

The displacement for part B is

$\begin{array}{rcl}\overline{{V}_{b}}& =& \frac{{X}_{b}}{{t}_{b}}\\ {X}_{b}& =& 15.6\times 1.75\\ {X}_{b}& =& 27.3\u200am\\ & & \end{array}$

The displacement for part is $27.3$meter.

The total displacement will be $29.9+27.3\text{}=\text{}57.2$meter.

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