Question

A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is$\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{6}\hat{\mathbf{i}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{1}\hat{\mathbf{j}}\mathbf{)}\mathit{m}\mathbf{/}\mathit{s}$, with$\hat{\mathbf{i}}$ horizontal and$\hat{j}$ upward. (a)To what maximum height does the ball rise? (b)What total horizontal distance does the ball travel? What are the (a) magnitude and (d) angle (below the horizontal) of the ball’s velocity just before it hits the ground?

Short answer

(a) Maximum height attained by the ball is 11.0m

(b) Total horizontal distance traveled by the ball is 23m

(c) Magnitude of the ball’s velocity when it is about to hit the ground is 17 m/s

(d) Angle of the ball’s velocity when it is about to the hit the ground is$-63°$

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} y=9.1m \\ \overrightarrow{v}=7.6\hat{i}+6.1\hat{j} \end{gathered}$$

Determining the concept

This problem is based on kinematic equations of motion that describe the motion of an object with constant acceleration. Using the velocity vector of the ball given at the distance, the initial velocity and its angle can be found. Using this, find the rest of the values.

Formulae:

The final velocity in kinematic equation can be written as,

$v_f^2=v_0^2+2ay$ (i)

$v_f=v_0+at$ (ii)

The magnitude of the velocity is,

$v=\sqrt{v_x^2+v_y^2}$ (iii)

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$ (iv)

(a) Determining the maximum height attained by the ball

For the velocity in y direction the equation (i) can be written as,

$$\begin{gathered} v_{fy}^2=v_{0y}^2+2ay\left(v\right) \\ {\left(6.1\right)}^2=v_{0y}^2-\left(2\times 9.8\times 9.1\right) \\ 37.21=v_{0y}^2-178.36 \\ {v}_{0y}^2=37.21+178.36 \\ {v}_{0y}^2=215.57 \\ {v}_{0y}^2=\sqrt{215.57} \\ {v}_{0y}^2=14.7m/s \end{gathered}$$

As there is no acceleration in the horizontal direction, so the velocity will be constant throughout the motion.

$v_{0x}=7.6m/s$

At the maximum height of the projectile,$v_{fy}=0$ , thus the equation (v) can be written as,

$$\begin{gathered} 0={\left(14.7\right)}^2-\left(2\times 9.8\times y_{max}\right) \\ {y}_{max}=\frac{14.7^2}{2\times 9.8} \\ {y}_{max}=11.0m \end{gathered}$$

(b) Determining the total horizontal distance travelled by the ball

Time required to reach the maximum height is given by equation (ii) and can be written as,,

$v_{fy}=v_{0y}+at$

For maximum height, $v_{fy}=0$

$$\begin{gathered} 0=14.7-9.8\times t \\ \frac{14.7}{9.8}=t \\ t=1.5s \end{gathered}$$

Time required for total flight,

role="math" localid="1661146549954" $$\begin{gathered} T=2\times timerequiredtoreachthemaximumheight \\ =2\times 1.5 \\ T=3.0sec \end{gathered}$$

Total Horizontal distance travelled by the ball,

$$\begin{gathered} speed=\frac{dis\tan ce}{time} \\ dis\tan ce=speed\times time \\ =7.6\times 3 \end{gathered}$$

$$\begin{gathered} dis\tan ce=22.8 \\ \approx 23m \end{gathered}$$

(c) Determining the magnitude of the ball’s velocity when it is about to hit the ground

When the ball is about to hit the ground, its velocity components will be 7.6 m/s in the x direction and -14.7 m/s in the y direction. The magnitude of the velocity is then given by equation (iii)

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ =\sqrt{7.6^2+{\left(-14.7\right)}^2} \\ =\sqrt{57.76+216.09} \\ v=16.5 \\ \approx 17m/s \end{gathered}$$

(d) Determining the angle of the ball’s velocity when it is about to the hit the ground

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \\ ={\tan }^{-1}\left(-\frac{14.7}{7.6}\right) \\ \theta =-62.66° \\ \approx -63° \end{gathered}$$

With the sign of the components, it is said that the resultant of the vector lie in the fourth quadrant. Therefore, the maximum height attained, total horizontal distance travelled by the ball, angle of the ball’s velocity when it is about to the hit the ground can be found using the kinematic equations and concept of projectile motion.