Question

A football player punts the ball at a $\mathbf{45}.\mathbf{0}$angle. Without an effect from the wind, the ball would travel $\mathbf{60}.\mathbf{0}\mathbf{m}$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 $\mathbf{1}.\mathbf{50}\mathbf{m}/\mathbf{s}$. What distance does the ball travel horizontally?

Short answer

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

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

Step-by-step solution

Definition of initial velocity 

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.

given data

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/s²

The initial velocity we have to calculate.

The initial velocity of the ball

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

$\begin{array}{rcl}{R}_x& =& \frac{{V_i}^{2}sin2{\theta }_i}{g}\\ 60& =& \frac{{V_i}^{2}sin2\left(45\right)}{9.8}\\ V_i& =& \sqrt{588}\\ V_i& =& 24.2 m/s\\ & & \end{array}$

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

Distance travelled by the ball

The Components of the initial velocity.

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

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

The velocity in the direction for frame B will be

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

Time is needed for finding the distance

The Y- Components of the initial velocity.

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

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

Putting the value of the given data in the equation

$\begin{array}{rcl}{V}_f& =& V_i+at\\ -17.1& =& 17.1+\left(-9.8\right)t\\ t& =& 3.49 s\\ & & \end{array}$

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$t_a=\frac{3.49}{2}=1.75 s$

The displacement for part A is

$\begin{array}{rcl}\overline{V_a}& =& \frac{X_a}{t_a}\\ X_a& =& 17.1\times 1.75\\ X_a& =& 29.9 m\\ & & \end{array}$

The displacement for part A is $29.9$meter

The displacement for part B is

$\begin{array}{rcl}\overline{V_b}& =& \frac{X_b}{t_b}\\ X_b& =& 15.6\times 1.75\\ X_b& =& 27.3 m\\ & & \end{array}$

The displacement for part is $27.3$meter.

The total displacement will be $29.9+27.3=57.2$meter.