SOLUTION: use the information s(t)= -16t^2 + v0t + s0
We throw a rock into the air with initial velocity of 50 ft/sec, and initial position of 6 ft. Find how high the rock goes before
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Question 189344: use the information s(t)= -16t^2 + v0t + s0
We throw a rock into the air with initial velocity of 50 ft/sec, and initial position of 6 ft. Find how high the rock goes before coming back down. To answer this question, fill in the blanks in the following sentence, where the first blank gives this height, and the second blank gives how many seconds into the flight of the rock it reaches this maximum height.
The rock reaches its highest position of _______ feet above the ground after _______ seconds of flight
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
use the information s(t)= -16t^2 + v0t + s0
We throw a rock into the air with initial velocity of 50 ft/sec, and initial position of 6 ft. Find how high the rock goes before coming back down. To answer this question, fill in the blanks in the following sentence, where the first blank gives this height, and the second blank gives how many seconds into the flight of the rock it reaches this maximum height.
The rock reaches its highest position of _______ feet above the ground after _______ seconds of flight
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v0 = 50 ft/sec
s0 = 6 feet
s(t) = -16t^2 + 50t + 6
The easiest way (without calculus) to do these is to set s(t) = 6, and solve for the times when the object is at 6 feet. One will be zero, the other will be on the way down. The average of the 2 times will be the time the object was at its max height.
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-16t^2 + 50t + 6 = 6
16t^2 = 50t
t = 0
t = 50/16 seconds
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The average, or midpoint, is 25/16 seconds.
s(25/16) = -16*(25/16)^2 + 50*(25/16) + 6
max = -625/16 + 1250/16 + 6
max = 721/16 feet, or 45 ft 3/4 inches
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