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) = -16t2 + 22 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? -H+22 ft/sec What should its initial velocity be in order to strike the ground after 1 second? -232.2 ft/sec 5 seconds? ft/sec T seconds? ft/sec

Explanation / Answer

for H feet ans = root{2*32*(H-224)} displacement = -224 ft, therefore u=-208ft/sec for 5 sec := 35.2 ft/sec T sec:= {-224+16*(T^2)}/T

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.