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4. Please solve the question. Thanks!!

Mathematics
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Apr 7th, 2015

h = -32x^2/4900 +x+190

dh/dx = -64x/4900 +1

for max height let dh/dx = 0  hence we get, -64x/4900 =-1    or x = 4900/64 =76.5625  (answer A)

H(76.5625) = -32  (76.5625) ^2/4900 +76.5625+190 =228.2813 Feet   (answer B)

(c)    the projectile was projected from an height of 190 ft (x = 0)  . hence when it will touch the water ,height H will become -190 therefore   -32x^2/4900 +x+190 =   -190

solving this we get x = 329.64 feet   (Answer C)

(D)  Correct Graph : B



Apr 7th, 2015

When the height of the projectile is 100 feet above the water, how far is it from the cliff? Do I just plug in 100 for x?

Apr 7th, 2015

x = 216.72  when H = 100 ft

Dear Answer of C above is x = 263.53 ft     not 329.64

Apr 7th, 2015

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Apr 7th, 2015
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