Question

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure, where${}^{\mathit{t}\mathbf{=}\mathbf{0}}$at the instant the ball is struck. The scaling on the vertical axis is set by${}^{{\mathbf{v}}_{\mathbf{a}}\mathbf{=}\mathbf{19}\mathbf{}\mathbf{m}\mathbf{/}\mathit{s}}$and${}^{{\mathit{v}}_{\mathbf{b}}\mathbf{=}\mathbf{31}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$. (a) How far does the golf ball travel horizontally before returning to ground level? (b) What is the maximum height above ground level attained by the ball?

Short answer

(a) The horizontal distance travelled by the golf ball is${}^{95\mathrm{m}}$

(b) The maximum height above the ground level attained by the ball is${}^{31\mathrm{m}}$

Step-by-step solution

Given information

$$\begin{gathered} {\mathrm{v}}_{\mathrm{a}}=19\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{b}}=31\mathrm{m}/\mathrm{s} \\ \mathrm{t}=5\sec \end{gathered}$$

Determining the concept of kinematic equation

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Using these equations, the maximum height attained by the ball and the horizontal distance travelled by it can be calculated.

Formula:

The Newton’s second kinematic equation is given by,

$∆y=\left(\begin{array}{c}{\mathrm{v}}_{0\mathrm{y}}\end{array}\right)\left(\begin{array}{c}\mathrm{t}\end{array}\right)+\frac{1}{2}\times a+t^2$ (i)

The horizontal displacement of the projectile is

$$\begin{gathered} x=v_{0\mathrm{x}}\times t \\ =v_{0}cos\theta \times t \end{gathered}$$

(ii)

(a) Determining the horizontal distance travelled by the golf ball

The graph given in the figure is plotted as,${}^{v=\sqrt{v_x^2+v_y^2}}$ vs ${}^t$.

By observing the graph, at${}^{t=2.5s}$, the velocity of the ball is${}^{19\mathrm{m}/\mathrm{s}}$means${}^{v_y=0}$and${}^{v_x=19m/s}$as the ball reaches the maximum height.

In 5 seconds the ball returns to the ground. The distance travelled by the golf ball before returning to the ground is given by equation (ii),

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{0\mathrm{x}}\times \mathrm{t} \\ \mathrm{x}=19\times 5 \\ =95\mathrm{m} \end{gathered}$$

(b) Determining themaximum height above the ground level attained by the ball

The initial velocity of the ball is${}^{{\mathrm{v}}_0=31\mathrm{m}/s}$

As ${}^{v_0=\sqrt{v_{0x}^2+v_{0y}^2}}$and${}^{{\mathrm{v}}_{0\mathrm{x}}=19\mathrm{m}/s}$

$$\begin{gathered} 31=\sqrt{{19}^2+{\mathrm{v}}_{0\mathrm{y}}^2} \\ {\mathrm{v}}_{0\mathrm{y}}=24.5\mathrm{m}/s \end{gathered}$$

Atlocalid="1654237884667" ${}^{t=2.5s}$, the maximum height reached by the ball is given by equation (i),

$$\begin{gathered} ∆\mathrm{y}=\left(\begin{array}{c}{\mathrm{v}}_{0\mathrm{y}}\end{array}\right)\left(\begin{array}{c}\mathrm{t}\end{array}\right)-\frac{1}{2}{\mathrm{gt}}^2 \\ ∆\mathrm{y}=24.0\left(\begin{array}{c}2.5\end{array}\right)-\frac{1}{2}9.8{\left(\begin{array}{c}2.5\end{array}\right)}^2 \\ ∆\mathrm{y}=30.63 \\ ~31\mathrm{m} \end{gathered}$$

The maximum height above the ground level attained by the ball is${}^{31\mathrm{m}}$