Problema Solution
a rock is thrown upwards from a height of 200 feet with an initial velocity of 45 feet/second. the height of the rock above the ground t seconds after being released is given by h(t)=-16t^2+45t+200
match the following
h(2.5)
h(1.5)
h(1.0)
h(0.5)
h(2.0)
Answer provided by our tutors
h(2.5) = -16(2.5^2)+45*2.5+200
h(2.5) = 212.5
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h(1.5) = -16(1.5^2)+45*1.5+200
h(1.5) = 231.5
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h(1.0) = -16(1.0^2)+45*1.0+200
h(1.0) = 229
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h(0.5) = -16(0.5^2)+45*0.5+200
h(0.5) = 218.5
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h(2.0) = -16(2.0^2)+45*2.0+200
h(2.0) = 226
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