Question

A ball is kicked on Earth from a location$<9,0,-5>$ (on the ground) with initial velocity role="math" localid="1656668041027" $\mathbf{<}\mathbf{-}\mathbf{10}\mathbf{,}\mathbf{13}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{>}\mathbf{m}\mathbf{/}\mathbf{s}$ . Neglecting air resistance: (a) What is the velocity of the ball 0.6 s after being kicked? (b) What is the location of the ball 0.6 s after being kicked? (c) What is the maximum height reached by the ball?

Short answer

a) The velocity of the balls after the time of 0.6 s is .

b) The location of the balls after the time of 0.6 s is role="math" localid="1656668244576" .

c) The value of maximum height. of the ball is $8.622\mathrm{m}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The location of the point from where the ball has been kicked is
• The initial velocity of the ball when it was kicked is

Explanation of the velocity and first and third equation of motion

The velocity of the car has two crucial factors that are magnitude and direction, and hence it is the vector quantity. The time and displacement are the dependent factors of the velocity.

The relation of the final and initial velocity of an object with the time used to obtain the value of acceleration is explained by the first equation of motion. It is expressed as follows:

$v=u+at$

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

In the effect of uniform acceleration and knowing the value of initial velocity and distance traveled by a specific item, the final velocity can be determined using the thirst equation of motion. It is expressed as follows:

$v^2=u^2+2as$

Here, s is the distance.

Determination of the velocity of the balls after the time of 0.6 s

(a)

Write the expression for the velocity of the ball in the y-direction as the motion is vertical.

Take the y component from the coordinate point of velocity and use the first equation of motion.

$$\begin{gathered} \mathrm{v}=13-\left(9.8\times 0.6\right) \\ =7.12\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the balls after the time of 0.6 s is .

Determination of the location of the balls after the time of 0.6 s

(b)

Determine the average velocity of the ball by finding the average of the y-component in the coordinate point of the velocity.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{avg}}=\left(-10,\frac{13+7.12}{2},-5\right)\mathrm{m}/\mathrm{s} \\ =\left(-10,10.06,-5\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

Determine the distance by multiplying the time in the coordinate point of the velocity.

Determine the location of the ball by adding the above distance in the coordinate point of location.

Thus, the location of the balls after the time of 0.6 s is .

Determination of the maximum height attained by the ball

(c)

It is known that at the maximum height, the ball's final velocity will be zero. So, use the third equation of motion and determine the value of maximum height.

$0=u^2+2as$

Here, a is the acceleration which is equal to the acceleration due to gravity in the upward direction, that is, $-g=-9.8m/s^2$.

Substitute all the values in the above expression.

$$\begin{gathered} 0=u^2+2\left(-g\right)s \\ {u}^2=2gs \\ s=\frac{u^2}{2g} \\ =\frac{{\left(13m/s\right)}^2}{2\left(9.8m/s^2\right)} \\ =8.622m \end{gathered}$$

Thus, the value of the maximum height of the ball is $8.622\mathrm{m}$.