Question

The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was 45° and neglecting air resistance; determine (a) the initial speed of the projectile and (b) the total time interval the projectile was in flight. (c) How would the answers change if the range were the same but the launch angle were greater than 45°? Explain.

Short answer

1. The initial speed of the projectile is 28.19 m/s .
2. The total time interval is 4.068 s.
3. Solution to this part is explained later on in the solution

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The record distance is, $x=81.1 \text{m}$.

The initial launch angle is, $\theta ={45}^0$.

Finding the equation for time

Since the ground is levelled, final velocity in y direction is equal to its initial velocity but in opposite direction.

i.e. ${\mathit{v}}_{\mathbf{y}}\mathbf{=}\mathbf{-}{\mathit{u}}_{\mathbf{y}}$

Hence,

role="math" localid="1663715404450" $\begin{array}{rcl}{\mathit{v}}_{\mathbf{y}}& \mathbf{=}& {\mathit{u}}_{\mathbf{y}}\mathbf{+}{\mathit{a}}_{\mathbf{y}}\mathit{t}\\ \mathbf{-}{\mathit{u}}_{\mathbf{y}}& \mathbf{=}& {\mathit{u}}_{\mathbf{y}}\mathbf{-}\mathit{g}\mathit{t}\\ \mathit{g}\mathit{t}& \mathbf{=}& \mathbf{2}{\mathit{u}}_{\mathbf{y}}\\ \mathit{t}& \mathbf{=}& \frac{\mathbf{2}{\mathbf{u}}_{\mathbf{y}}}{\mathbf{g}}\end{array}$

Take the horizontal and vertical components for the initial velocity,

$$\begin{gathered} {\mathit{u}}_{\mathbf{x}}\mathbf{=}\mathit{u}\mathit{c}\mathit{o}\mathit{s}\mathbf{45} \\ {\mathit{u}}_{\mathbf{y}}\mathbf{=}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathbf{45} \end{gathered}$$

So,

$\mathit{t}\mathbf{=}\frac{\mathbf{2}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{45}}{\mathbf{g}}$

Solving for initial velocity (a)

Since we know the range,

$$\begin{gathered} x=u_{x}t \\ x=\left(u\cos \theta \right)\frac{2u\sin \theta }{g} \\ x=\frac{u^2}{g}\left(2\sin \theta \cos \theta \right) \\ x=\frac{u^2}{g}\left(\sin 2\theta \right) \end{gathered}$$

Substitute all the value in the above equation.

role="math" localid="1663715699914" $\begin{array}{rcl}81.1 \text{m}& =& \frac{u^2}{9.8 {\text{m/s}}^2}\left(\sin 2\times {45}^0\right)\\ u& =& 28.19 \text{m/s}\\ & & \end{array}$

Hence the initial speed of the projectile is 28.19 m/s.

Solving for total time interval in flight (b)

The expression of the total time interval in flight is expressed as,

$t=\frac{2u\sin \theta }{g}$

Substitute all the value in the above equation.

$$\begin{gathered} t=\frac{2(28.19 \text{m/s})\left(\sin {45}^0\right)}{9.8 {\text{m/s}}^2} \\ t=4.068 \text{s} \end{gathered}$$

Hence the total time interval is 4.068 s.

Explanation for (c):

Take the equation for we just formulated.

$x=\frac{u^2}{g}\left(\sin 2\theta \right)$

$$\begin{gathered} u^2=\frac{xg}{\sin 2\theta } \\ u=\sqrt{\frac{xg}{\sin 2\theta }} \end{gathered}$$

Hence, if we keep the range (x) constant and change the angle $\theta$, $\sin 2\theta$will increase for $\theta <90°$.

Therefore, to keep the range constant at a higher angle$(\theta >45°)$ , we will need to increase the initial velocity.

Similarly in the time equation, we see that with increasing initial velocity, its flight time also increases.

Hence, if $(\theta >45°)$, initial velocity and time both increases.