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, 0≤x≤350 .

Question

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,  0≤x≤350  .

Caren · Accepted Answer

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.&quot;. Given,h(x)=−32x21002+xFind 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 &quot;x&quot; we get,h′(x)=ddx[−32x21002+x]h′(x)=ddx−32x210000+ddx(x)Use the power rule of differentiation we get,ddxxn=nxn−1h′(x)=−32(2x)10000+1h′(x)=−64x10000+1Now Substitute h′(x)=0 we get,−64x10000+1=064x10000=164x=10000x=1000064x=156.25If f″(x)&amp;gt;0 then function attains minimum value at x=aIf f″(x)&amp;lt;0 then function attains minimum value at x=a.Now-Again differentiate the function we get,h″(x)=ddx−64x10000−1h″(x)=−6410000=0h″(x)=−156.25&amp;lt;0h″(156.25)=−156.25&amp;lt;0Therefore the function attains a maximum value at x=156.25Therefore. The horizontal distance from the firing point is the height of the projectile a maximum is 156.25Step 3b) 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 78.125Step 4C) 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=0x−

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

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