A 3 kg flowerpot drops from a tall building. The initial speed of the pot is zer

Question

A 3 kg flowerpot drops from a tall building. The initial speed of the pot is zero, and you may neglect air resistance.

a) What is the magnitude of the force acting on the pot while it is in the air 1.5 s after it begins to fall?

b) After the pot has fallen 26 m, what is its speed?

c) After the pot has fallen 26 m, it enters a viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?

d) What is the force exerted on the liquid by the pot?

Explanation / Answer

F - mg = m a >>> F = m (g + a) = 3 * (9.8 + 169.86) = 538.98 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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