A blue ball is thrown upward with an initial speed of 20.2 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 20.2 m/s, from a height of 0.8 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 22.8 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. How long after the blue ball is thrown are the two balls in the air at the same height? I got 0.8208 but my feedback was "that answer would be correct if both balls were thrown at the same time. The problem states the red ball was thrown downward 2.5 seconds after the blue ball was thrown. One solution is to determine the position and velocity of the blue ball at the time the red ball was thrown, and then use those as the initial condition."

Explanation / Answer

let the time be t s of blue=ut+0.5*at2=20.2*t-4.905*t^2 s of red=6.6*(t-2.5)+ 4.905*(t-2.5)^2 sum of the distances=22.8 20.2t+2.5^2*4.905-24.525t+6.6t-16.5=22.8 t=3.8sec 6)plot B is correct 7)the acceleration is positive and it is speeding up

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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