A football player punts the football so that it will have a “hang time” (time of flight) of $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}$ and land $\mathbf{46}\mathbf{}\mathbf{m}$ away. If the ball leaves the player’s foot 150cm above the ground, what must be the (a) magnitude and (b) angle (relative to the horizontal) of the ball’s initial velocity?

It is given that,

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

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,

${\mathrm{v}}_0=\frac{\mathrm{x}}{\mathrm{t}}$ (i)

$\mathrm{y}={\mathrm{v}}_{\mathrm{oy}}\mathrm{t}-\frac{1}{2}{\mathrm{at}}^2$ (ii)

where, yis total displacement, is an initial velocity, t is time and a is an acceleration.

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,

role="math" localid="1657364079185" $$\begin{gathered} {\mathrm{V}}_{0\mathrm{x}}=-\frac{46}{4.5} \\ =10.22\mathrm{s} \end{gathered}$$

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

$\mathrm{y}={\mathrm{v}}_{0\mathrm{y}}\mathrm{t}+\frac{1}{1}{\mathrm{at}}^2$

$-1.5={\mathrm{V}}_{0\mathrm{y}}4.5-0.5\times 9.8\times 4.5^2$

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

role="math" localid="1657364149809" $$\begin{gathered} {\mathrm{V}}_{0\mathrm{y}}=\frac{-1.5+\left(0.5\times 9.8\times 4.5^2\right)}{4.5} \\ {\mathrm{V}}_{0\mathrm{y}}=21.72\mathrm{m}/\mathrm{s} \end{gathered}$$

So, magnitude of initial velocity is,

role="math" localid="1657364179050" $$\begin{gathered} \mathrm{V}=\sqrt{10.{22}^2+21.{72}^2} \\ \mathrm{V}=24\mathrm{m}/\mathrm{s} \end{gathered}$$

So, magnitude of initial velocity is$24\mathrm{m}/\mathrm{s}$

The angle is,

$$\begin{gathered} \mathrm{tan\theta }=\frac{21.72}{10.22} \\ \theta =64.{80}^{°} \\ \approx {65}^{°} \end{gathered}$$

Therefore, the angle is${65}^{°}$