SOLUTION: A rock is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. The rock travels up to a maximum height and returns downward, just mi
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-> SOLUTION: A rock is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. The rock travels up to a maximum height and returns downward, just mi
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Question 1089544: A rock is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. The rock travels up to a maximum height and returns downward, just missing the building and striking the ground below.
(a) Give the function (s(t)) that describes the height of the rock in terms of time t.
(b) Determine the time at which the rock reaches its maximum height and the maximum height in
feet.
(c) For what time interval will the rock be more than 300 ft above ground level? (Round your
answer to two decimal places.)
(d) After how many seconds will it hit the ground? (Round your answer to two decimal places.)
(e) Sketch a graph of s(t) for t > 0 and label the points on the graph that correspond to the answers
found in parts (1b),(1c), and (1d). Label the points (1b),(1c), and (1d). Answer by ikleyn(52781) (Show Source):
Your basic function for the height above the ground as the function of time is
h(t) = .
MEMORIZE : a) the number 16 is the half of the gravity acceleration g = 32 ft/s^2; the first term -16t^2 goes with the sign "-" (minus).
b) the vertical velocity stands as the coefficient at t (200) (positive if the initial velocity is directed up);
c) the initial height stands as the constant term (50).
