An object is dropped from a tower of height 224 feet, subject to only the consta

Question

An object is dropped from a tower of height 224 feet, subject to only the constant acceleration of -32 ft/sec2 due to gravity. Write the equation for the height of the object from the ground at time t if the object is dropped from the tower. s(t) = For the situation in part (a), how many seconds pass before the object strikes the ground? sec What is its velocity at that instant? ft/sec Suppose the object is instead thrown upward. What should its initial velocity be in order to reach a maximum height of 300 feet? ft/sec 400 feet? ft/sec H feet? ft/sec What should its initial velocity be in order to strike the ground after 1 second? ft/sec 5 seconds? ft/sec T seconds? ft/sec

Explanation / Answer

(t) = -32 ft/sec v(t) = Integral [a(t) dt] = -32t + C v(0) = 0 [Initial Velocity] v(0) = -32(0) + C C = 0 so v(t) = -32t s(t) = Integral [v(t) dt] = -32t^2 / 2 + C s(0) = 384 [Initial Height] s(0) = -32(0)^2 /2 + C C = 384 s(t) = -32t^2 / 2 + 384 = -16t^2 + 384 So the height of the object at time t can be expressed as s(t) = -16t^2 + 384

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