Analyzing the Motion of a Projectile A projectile is fired at an inclinatio

FobelloE

FobelloE

Answered question

2021-10-07

Analyzing the Motion of a Projectile A projectile is fired
at an inclination of 45 to the horizontal, with a muzzle
velocity of 100 feet per second. The height h of the projectile
is modeled by h(x)=32x21002+x
where x is the horizontal distance of the projectile from the firing point
(a) At what horizontal distance from the firing point is the height of the projectile a maximum?
(b) Find the maximum height of the projectile.
(c) At what horizontal distance from the firing point will the projectile strike the ground?
(d) Using a graphing utility, graph the function h, 0x350 .

Answer & Explanation

Caren

Caren

Skilled2021-10-08Added 96 answers

Since you have posted a question with multiple subparts, we will solve the first three subparts for you. To get the remaining subparts solved please repost the complete question and mention the subparts to be solved.". Given,
h(x)=32x21002+x
Find the Critical values of the above function:
To find the critical value find the first derivative of the function and equate it to zero,
Now differentiate the above function with respect to "x" we get,
h(x)=ddx[32x21002+x]
h(x)=ddx32x210000+ddx(x)

Use the power rule of differentiation we get,
ddxxn=nxn1
h(x)=32(2x)10000+1
h(x)=64x10000+1
Now Substitute h(x)=0 we get,
64x10000+1=0
64x10000=1
64x=10000
x=1000064
x=156.25
If f(x)>0 then function attains minimum value at x=a
If f(x)<0 then function attains minimum value at x=a.
Now-Again differentiate the function we get,
h(x)=ddx64x100001
h(x)=6410000=0
h(x)=156.25<0
h(156.25)=156.25<0
Therefore the function attains a maximum value at x=156.25
Therefore. The horizontal distance from the firing point is the height of the projectile a maximum is 156.25
Step 3
b) Find the maximum height of the projection:
Now Substitute x=156.25 in the given equation we get,
h(156.25)=32(156.25){2}156.25+156.25
=32(24414.0625)10000+156.25
=78125010000+156.25
=78.125+156.25
=78.125 Therefore the maximum height of the projectile is 78.125
Step 4
C) At what horizontal distance from the firing point will the projectile strike the ground:
To find the x-intercepts:
Put h(x)=0 we get,
32x21002+x=0
x

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