A blue ball is thrown upward with an initial speed of 21.6 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 21.6 m/s, from a height of 0.5 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 11.1 m/s from a height of 26.7 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. 3) What is the maximum height the blue ball reaches? 5) How long after the blue ball is thrown are the two balls in the air at the same height? could you show me how to get the answers please.!!

Explanation / Answer

Time taken for B ball to reach top: 21.6/9.81 = 2.2 sec The top = 0.5 + 21.6*2.2 - 0.5*9.81*2.2^2 = 24.3 meter When R ball is thrown, B ball has traveled 1/2*g(2.6-2.2)^2 from its top, or 0.5*9.81 *(2.6-2.2)^2 = 0.78 meters (from its top), or (24.3-0.78) = 23.52 meters from the ground. The moment B ball reaches 23.52 meters from the ground, R ball is thrown. When B ball and R ball have the same height (h-B = h-R): height of B ball = 23.52 - 0.5*g*t^2 height of R ball =11.1*t + 0.5*g*t^2 solve this equation with WolframAlpha (http://www.wolframalpha.com/input/i=-23.52+11.1*t+9.81*t^2), we get: t = 1.08 secs For the time since blue ball was thrown: 1.08 + 2.6 = 3.68 secs

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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