A mountain climber stands at the top of 53.4m cliff that overhangs a calm pool o

Question

A mountain climber stands at the top of 53.4m cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.79s apart and observes that they cause a single splash. The first stone has an initial velocity of -1.84m/s. How long after release of the first stone will the two stones hit the water? The acceleration of gravity is 9.8m/s^2. Answer in units of s, what initial velocity must The second stone have if they are to hit simultaneously, answer in units of m/s. what will be the velocity of the second stone the instant both stones hit the water Answer in units of m/s.

Explanation / Answer

dist. = v * t + 1/2 * g * t^2 describes the motion of a stone with initial speed. Replacing with numerical values 56.8 = 2.37 * t + 1/2 9.8 m/s * t^2 this is a good old quadratic: 2.37 * t + 1/2 9.8 m/s * t ^2 - 56.8 = 0 t = 3.1714 seconds is the time for the first stone. Now the second stone is late. it has to be there in 3.17 - 1.38 = 1.79 s again with dist. = v * t + 1/2 * g * t^2, but this time we do not know v (56.8 - 0. 5 * 9.8 * 1.79 ^2) / 1.79 = 22.96 m / s as initial speed the second stone has a final speed of: v + a * t = 22.96 + 9.8 * 1.79 = 40.5 m/sec or 145 km/h

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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