Respuesta :
Answer:
Using the equation given, we have to derive, because the derivative of a displacement functions results in velocity, and then we just calculate time, like this:
[tex]s=-16t^{2}+v_{0}t+s_{0}[/tex]
We know that [tex]s_{o}=0[/tex] and [tex]v_{0}=128ft/sec[/tex]
Replacing these values: [tex]s=-16t^{2} +128t+0[/tex]
Then, we derive: [tex]v=2(-16)t^{2-1} +128x^{1-1}\\ v=-32t+128t^{0} \\v=-32t+128[/tex]
But, when the object reach the highest point, the velocity in that moment is zero. (v = 0)
So, [tex]0=-32t+128\\-128=-32t\\\frac{-128}{-32} =t\\t=4sec[/tex].
Therefore, the object takes 4 seconds long to reach the highest point. Also, this is a symmetrical movement (because is a quadratic function), this means that the object uses double of this time to reach the ground. So, after 8 seconds, the object will be back at ground level.
To solve the second part of the problem, we have to solve a quadratic inequality. A height less than 128 feet can be expressed like: [tex]s<128[/tex]
We know that [tex]s=-16t^{2} +128t[/tex]
Replacing this s expression on the inequality:
[tex]-16t^{2}+128t<128[/tex]
Solving this inequality:
[tex]-16t^{2} +128t-128<0\\[/tex]
Dividing by -16:
[tex]\frac{-16t^{2}+128t-128 }{-16}<0\\t^{2}-8t+8>0[/tex]; which is easier too solve.
Therefore, solving this quadratic equation, the roots are = 6.82843 and
=1.17157. So, the answer that makes sense is =1.17157, because the total time to reach the highest point is 4 and 6 seconds is more than 4.
Hence, at 1.2 seconds, the object will be in a position less than 128 feet.
After 8 seconds, the object will be back at ground level.
At 1.2 seconds, the object will be in a position less than 128 feet.
Given that,
The position equation s = [tex]-16t^{2} +v_o+s_0[/tex]
where s represents the height of an object (in feet),
[tex]v_0[/tex] represents the initial velocity of the object (in feet per second),
[tex]s_0[/tex] represents the initial height of the object (in feet),
and t represents the time (in seconds).
initial velocity of 128 feet per second.
According to the question,
- Using the equation given, we have to derive, because the derivative of a displacement functions results in velocity, and then we just calculate time, like this:
[tex]s = -16t^{2} +v_o+s_0[/tex]
Where [tex]s_0 = 0 , v_0 = 128\frac{ft}{sec}[/tex]
Putting the given values,
[tex]s_0 = -16t^{2} +128t +0[/tex]
Differentiate both the sides,
[tex]v_0 = 2(-16)t^{2-1} +128\\v_0 = -32t +128[/tex]
When the object reach the highest point, the velocity in that moment is zero. (v = 0)
So, -32t + 128 = 0
32t = 128
t = [tex]\frac{128}{32}[/tex]
t = 4 sec
Therefore, the object takes 4 seconds long to reach the highest point. Also, this is a symmetrical movement (because is a quadratic function), this means that the object uses double of this time to reach the ground. So, after 8 seconds, the object will be back at ground level.
- To solve the second part of the problem, we have to solve a quadratic inequality. A height less than 128 feet can be expressed like:
[tex]s = -16t^{2} + 128[/tex]
Replacing the inequality
[tex]-16t^{2} + 128t < 128\\-16t^{2} + 128t - 128 < 0[/tex]
Divided by 16 both the sides,
[tex]x^{2} -8t+8 > 0[/tex]
Therefore, solving this quadratic equation, the roots are = 6.82843 and 1.17157. So, the answer that makes sense is = 1.17157, because the total time to reach the highest point is 4 and 6 seconds is more than 4.
Hence, at 1.2 seconds, the object will be in a position less than 128 feet.
For the more information about the Average Velocity click the link given below.
https://brainly.com/question/181536