Question 10

A ball is tossed into the air on the planet Ork. The height of the ball in meters above the ground, [tex]\( t \)[/tex] seconds after being thrown, is given by:

[tex]\[ s(t) = 8t^2 - 3t^3 \][/tex]

If necessary, round results to at least 3 decimal places accuracy.

a. Find the velocity and acceleration functions.

[tex]\[
\begin{array}{l}
v(t) = \square \\
a(t) = \square
\end{array}
\][/tex]

b. Over what interval(s) of time is the ball moving in the positive (upward) direction? (Give result in interval notation.)

Upward for [tex]\( t \in \square \)[/tex]

c. After initially being thrown in the air, when does the ball return to the ground? (Hint: What is the ball's position when it returns to the ground?)

The ball returns to the ground at time [tex]\( t = \square \)[/tex] seconds

d. What is the velocity of the ball when it returns to the ground?

[tex]\[ \square \, \text{m/sec} \][/tex]



Answer :

Alright, let’s solve each part of the problem step by step.

### Given:
The height of the ball as a function of time [tex]\( t \)[/tex] in seconds is:
[tex]\[ s(t) = 8t^2 - 3t^3 \][/tex]

### Part (a): Find the velocity and acceleration functions.
To find the velocity, we need to differentiate [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex].

[tex]\[ v(t) = \frac{d}{dt}[8t^2 - 3t^3] \][/tex]

Calculating the derivative:
[tex]\[ v(t) = \frac{d}{dt} [8t^2] - \frac{d}{dt}[3t^3] \][/tex]
[tex]\[ v(t) = 16t - 9t^2 \][/tex]

To find the acceleration, we need to differentiate the velocity function [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex].

[tex]\[ a(t) = \frac{d}{dt} [16t - 9t^2] \][/tex]

Calculating the derivative:
[tex]\[ a(t) = 16 - 18t \][/tex]

Thus:
[tex]\[ v(t) = 16t - 9t^2 \][/tex]
[tex]\[ a(t) = 16 - 18t \][/tex]

### Part (b): Over what interval(s) of time is the ball moving in the positive (upward) direction?

The ball moves upwards when the velocity [tex]\( v(t) \)[/tex] is positive:
[tex]\[ v(t) > 0 \][/tex]

So, we solve the inequality:
[tex]\[ 16t - 9t^2 > 0 \][/tex]

Factoring the expression:
[tex]\[ t(16 - 9t) > 0 \][/tex]

Solving for the intervals where the product is positive:
[tex]\[ 0 < t < \frac{16}{9} \][/tex]

Thus, the ball is moving upwards for:
[tex]\[ t \in (0, \frac{16}{9}) \][/tex]

### Part (c): When does the ball return to the ground?

The ball returns to the ground when its height [tex]\( s(t) \)[/tex] is zero:
[tex]\[ s(t) = 0 \][/tex]

So, we solve the equation:
[tex]\[ 8t^2 - 3t^3 = 0 \][/tex]

Factoring out the common term:
[tex]\[ t^2(8 - 3t) = 0 \][/tex]

We get the solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ 8 - 3t = 0 \][/tex]
[tex]\[ t = \frac{8}{3} = 2.66666666666667 \][/tex] seconds (approximately)

Therefore, the ball returns to the ground at:
[tex]\[ t = 2.667 \][/tex] seconds (rounded to 3 decimal places)

### Part (d): What is the velocity of the ball when it returns to the ground?

To find the velocity when [tex]\( t = 2.667 \)[/tex]:

[tex]\[ v(2.667) = 16(2.667) - 9(2.667)^2 \][/tex]

Using the given result:
[tex]\[ v = -21.333 \][/tex] m/sec (rounded to 3 decimal places)

So, the velocity when the ball returns to the ground is:
[tex]\[ -21.333 \][/tex] m/sec

### Summary:
a. [tex]\[ \begin{array}{l} v(t) = 16t - 9t^2 \\ a(t) = 16 - 18t \end{array} \][/tex]
b. The ball moves upwards for [tex]\( t \in (0, \frac{16}{9}) \)[/tex].
c. The ball returns to the ground at [tex]\( t \approx 2.667 \)[/tex] seconds.
d. The velocity of the ball when it returns to the ground is [tex]\( -21.333 \)[/tex] m/sec.