In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult. Suppose a ball is spiked from a height of$\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{m}$with an initial speed oflocalid="1654581187572" ${}^{\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$at a downward angle oflocalid="1654581198810" ${}^{\mathbf{18}\mathbf{.}\mathbf{00}\mathbf{°}}$. How much farther on the opposite floor would it have landed if the downward angle were, instead,localid="1654581212383" ${}^{\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{°}}$?

Here acceleration is nothing but acceleration due to gravity. Thus take${}^{a=9.8m/s^2}$

Primary angle of velocity with vertical direction is${}^{18.0°}$

Primary angle of velocity with vertical direction is${}^{8.0°}$

$$\begin{gathered} v=20.0m/s \\ height=2.30m \end{gathered}$$

In this case${}^{y=h}$

This problem deals with the kinematic equations that describe the motion of an object at constant acceleration. With second kinematic equation, time required to the ball to reach the ground with both the given angles can be found. Further, with the calculated time distances will be calculated. By taking the difference between these distances the required answer can be found.

Formulae:

The displacement in kinematic equation can be written as,

$y=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Speed is given by,

$v=\frac{d}{t}$ (ii)

Speed in x direction is given by,

$$\begin{gathered} v_{0x}=20\sin \left(18.0\right) \\ =6.18m/s \end{gathered}$$

Similarly speed in y direction is,

$$\begin{gathered} v_{0x}=20\cos \left(18.0\right) \\ =19.02m/s \end{gathered}$$

Assuming downward as positive, equation (i) can be written as,

$$\begin{gathered} y=y_{0x}t+\frac{1}{2}a{t}^2 \\ 2.3=\left(6.18t\right)+\frac{1}{2}9.8t^2 \\ 2.3=6.18t+4.9t^2 \\ t=-1.56sort=0.3s \end{gathered}$$

Take the positive value of t, as time is not negative,

$t=0.3006s$

Using equation (ii) the distance in y direction will be,

$$\begin{gathered} d=v_{0y}\times t \\ 19.02\times 0.3006 \end{gathered}$$

Thus, primary distance is

$d_{primary}=5.72m$

The speed in x direction at angle${}^{8.0°}$is given by,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}=20\sin (8.0) \\ =2.78\mathrm{m}/\mathrm{s} \end{gathered}$$

The speed in y direction at angle${}^{8.0°}$is given by,

$$\begin{gathered} {\mathrm{v}}_{0y}=20\cos (8.0) \\ =19.80\mathrm{m}/\mathrm{s} \end{gathered}$$

Assuming downward as positive, equation (i) can be written as,

$$\begin{gathered} \mathrm{y}={\mathrm{y}}_{0\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ 2.3=\left(2.78\mathrm{t}\right)+\frac{1}{2}9.8{\mathrm{t}}^2 \\ 2.3=2.78\mathrm{t}+4.9{\mathrm{t}}^2 \\ \mathrm{t}=-1.02\mathrm{s}or\mathrm{t}=0.46\mathrm{s} \end{gathered}$$

Take the positive value of t, as time is not negative,

$t=0.45785s$

Thus,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{secondary}}={\mathrm{v}}_{0\mathrm{y}}\times \mathrm{t} \\ =20\cos \left(8.0\right)\times 0.45785 \\ {\mathrm{d}}_{\mathrm{secondary}}=9.07\mathrm{m} \\ ∆\mathrm{d}={\mathrm{d}}_{\mathrm{secondary}}-{\mathrm{d}}_{\mathrm{primary}} \\ =9.07-5.72 \\ =9.07-5.72 \\ \mathrm{Thus},∆\mathrm{d}=3.35\mathrm{m} \end{gathered}$$

Thus, the difference in the distances using the flight time and horizontal velocity calculated from the kinematic equations and vector resolution can be found.