Answer :
Sure, let's solve the given problem step by step:
### Part (a): At what time does the ball hit the ground?
The height of the ball above the ground is given by the function:
[tex]\[ s(t) = -16t^2 + 73 \][/tex]
To find the time at which the ball hits the ground, we need to determine when the height [tex]\( s(t) \)[/tex] equals zero:
[tex]\[ s(t) = 0 \][/tex]
[tex]\[ -16t^2 + 73 = 0 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 16t^2 = 73 \][/tex]
[tex]\[ t^2 = \frac{73}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{73}{16}} \][/tex]
[tex]\[ t = \frac{\sqrt{73}}{4} \][/tex]
Evaluating the expression:
[tex]\[ t \approx 2.136 \text{ seconds} \][/tex]
Thus, the ball hits the ground at approximately [tex]\( 2.136 \)[/tex] seconds.
---
### Part (b): What is the instantaneous velocity of the ball when it hits the ground?
The instantaneous velocity of the ball as a function of time is the derivative of the height function [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex].
Given:
[tex]\[ s(t) = -16t^2 + 73 \][/tex]
We find the derivative:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 73] \][/tex]
[tex]\[ v(t) = -32t \][/tex]
Now, we substitute the time when the ball hits the ground ( [tex]\( t = 2.136 \)[/tex] seconds ) into the velocity function:
[tex]\[ v(2.136) = -32 \times 2.136 \][/tex]
[tex]\[ v(2.136) \approx -68.352 \text{ feet per second} \][/tex]
Thus, the instantaneous velocity of the ball when it hits the ground is approximately [tex]\( -68.352 \)[/tex] feet per second.
---
### Part (c): What is the average velocity during its fall?
The average velocity during the fall can be calculated using the formula:
[tex]\[ \text{Average Velocity} = \frac{\text{Change in Displacement}}{\text{Total Time}} \][/tex]
Here, the initial height is [tex]\( 73 \)[/tex] feet, and the final height when the ball hits the ground is [tex]\( 0 \)[/tex] feet. The total time of fall is [tex]\( 2.136 \)[/tex] seconds.
Change in displacement:
[tex]\[ \Delta s = \text{Final Height} - \text{Initial Height} \][/tex]
[tex]\[ \Delta s = 0 - 73 \][/tex]
[tex]\[ \Delta s = -73 \text{ feet} \][/tex]
So, the average velocity is:
[tex]\[ \text{Average Velocity} = \frac{-73 \text{ feet}}{2.136 \text{ seconds}} \][/tex]
[tex]\[ \text{Average Velocity} \approx -4\sqrt{73} \approx -34.28 \text{ feet per second} \][/tex]
Thus, the average velocity during the ball's fall is approximately [tex]\( -4\sqrt{73} \text{ feet per second} \)[/tex].
To summarize:
(a) The ball hits the ground at [tex]\( \approx 2.136 \)[/tex] seconds.
(b) The instantaneous velocity when it hits the ground is [tex]\( \approx -68.352 \)[/tex] feet per second.
(c) The average velocity during the fall is [tex]\( \approx -4\sqrt{73} \text{ feet per second} \)[/tex].
### Part (a): At what time does the ball hit the ground?
The height of the ball above the ground is given by the function:
[tex]\[ s(t) = -16t^2 + 73 \][/tex]
To find the time at which the ball hits the ground, we need to determine when the height [tex]\( s(t) \)[/tex] equals zero:
[tex]\[ s(t) = 0 \][/tex]
[tex]\[ -16t^2 + 73 = 0 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 16t^2 = 73 \][/tex]
[tex]\[ t^2 = \frac{73}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{73}{16}} \][/tex]
[tex]\[ t = \frac{\sqrt{73}}{4} \][/tex]
Evaluating the expression:
[tex]\[ t \approx 2.136 \text{ seconds} \][/tex]
Thus, the ball hits the ground at approximately [tex]\( 2.136 \)[/tex] seconds.
---
### Part (b): What is the instantaneous velocity of the ball when it hits the ground?
The instantaneous velocity of the ball as a function of time is the derivative of the height function [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex].
Given:
[tex]\[ s(t) = -16t^2 + 73 \][/tex]
We find the derivative:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 73] \][/tex]
[tex]\[ v(t) = -32t \][/tex]
Now, we substitute the time when the ball hits the ground ( [tex]\( t = 2.136 \)[/tex] seconds ) into the velocity function:
[tex]\[ v(2.136) = -32 \times 2.136 \][/tex]
[tex]\[ v(2.136) \approx -68.352 \text{ feet per second} \][/tex]
Thus, the instantaneous velocity of the ball when it hits the ground is approximately [tex]\( -68.352 \)[/tex] feet per second.
---
### Part (c): What is the average velocity during its fall?
The average velocity during the fall can be calculated using the formula:
[tex]\[ \text{Average Velocity} = \frac{\text{Change in Displacement}}{\text{Total Time}} \][/tex]
Here, the initial height is [tex]\( 73 \)[/tex] feet, and the final height when the ball hits the ground is [tex]\( 0 \)[/tex] feet. The total time of fall is [tex]\( 2.136 \)[/tex] seconds.
Change in displacement:
[tex]\[ \Delta s = \text{Final Height} - \text{Initial Height} \][/tex]
[tex]\[ \Delta s = 0 - 73 \][/tex]
[tex]\[ \Delta s = -73 \text{ feet} \][/tex]
So, the average velocity is:
[tex]\[ \text{Average Velocity} = \frac{-73 \text{ feet}}{2.136 \text{ seconds}} \][/tex]
[tex]\[ \text{Average Velocity} \approx -4\sqrt{73} \approx -34.28 \text{ feet per second} \][/tex]
Thus, the average velocity during the ball's fall is approximately [tex]\( -4\sqrt{73} \text{ feet per second} \)[/tex].
To summarize:
(a) The ball hits the ground at [tex]\( \approx 2.136 \)[/tex] seconds.
(b) The instantaneous velocity when it hits the ground is [tex]\( \approx -68.352 \)[/tex] feet per second.
(c) The average velocity during the fall is [tex]\( \approx -4\sqrt{73} \text{ feet per second} \)[/tex].
