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
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?
x = 216.72 when H = 100 ft
Dear Answer of C above is x = 263.53 ft not 329.64
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