Q117P. - Fundamentals Of Physics

Question

A football player punts the football so that it will have a “hang time” (time of flight) of 4.5 s and land 46 m away. If the ball leaves the player’s foot 150 cm above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball’s initial velocity?

Step-by-Step Solution

Answer

(a) Magnitude of initial velocity is 24 m/s

(b)Therefore, the angle is $65°$

Step 1: Given information

It is given that,

$$\begin{gathered} x=46 \mathrm{m} \\ y=-1.5 \mathrm{m} \\ t=4.5 \mathrm{s} \end{gathered}$$

Step 2: Determining the concept

The problem is based on the standard kinematic equations that describe the motion of an object with constant acceleration. To find magnitude of initial velocity, use kinematic equations in both directions. One is in horizontal direction to find horizontal velocity and another is in vertical direction to find vertical velocity.

 

Formula:

 

The velocity and displacement is given by,

 $$\begin{gathered} v_0=\frac{x}{t} \\ y=v_{0y}t-\frac{1}{2}a{t}^2 \end{gathered}$$                                                                                                                          (i)

                                                                                                              (ii)

where, y is total displacement, $v_0$ is an initial velocity, t is time and a is an acceleration.

Step 3: (a) Determining the magnitude of initial velocity

With equation (i) the horizontal component of initial velocity is,

$V_{0x}=\frac{x}{t}$

As there is no acceleration in horizontal direction, usedirection formula for velocity,

$$\begin{gathered} V_{0x}=\frac{46}{4.5} \\ =10.22 \mathrm{s} \end{gathered}$$

Now, with equation (ii) vertical velocity component is,

$$\begin{gathered} y=v_{0y}t+\frac{1}{1}a{t}^2 \\ -1.5=v_{0y}4.5-0.5\times 9.8\times 4.5^2 \end{gathered}$$

Acceleration will be negative because velocity is upwards, which is positive, and if acceleration is in downwards , then velocity will be negative.

$$\begin{gathered} v_{0y}=\frac{-1.5+\left(0.5\times 9.8\times 4.5^2\right)}{4.5} \\ {v}_{0y}=21.72 \mathrm{m}/\mathrm{s} \\ \mathrm{So}, \mathrm{magnitude} \mathrm{of} \mathrm{initial} \mathrm{velocity} \mathrm{is}, \\ \mathrm{V}=\sqrt{10.{22}^2+21.{72}^2} \\ \mathrm{V}=24 \mathrm{m}/\mathrm{s} \end{gathered}$$

So, magnitude of initial velocity is 24 m/s

Step 4: (b) determining the angle

The angle is,

$$\begin{gathered} \tan \theta =\frac{21.72}{10.22} \\ \theta =64.80° \\ \approx 65° \\ \mathrm{Therefore}, \mathrm{the} \mathrm{angle} \mathrm{is} 65° \end{gathered}$$