Ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. after how long will the balls be at the same height?

Question

Yareli Joyce

Physics

15 February, 11:40

Ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. after how long will the balls be at the same height?

+4

Answers (1)

Know the Answer?

Waylon Blair · Answer 1

0.6 seconds after the balls start moving, they'll be at the same height. The altitude the 1st ball will be at time T is a = 25T - 0.5AT^2 The altitude the 2nd ball will be at time T is a = 15 - 0.5AT^2 Set both expressions equal to each other and solve for T 25T - 0.5AT^2 = 15 - 0.5AT^2 Add 0.5AT^2 to both sides 25T = 15 Divide both sides by 25 T = 15/25 = 3/5 = 0.6 So both balls will be at the same altitude 0.6 seconds after they start moving. Let's verify those results. Assume A = 9.8 m/s^2 0.5 * 9.8 m/s^2 * 0.6s^2 = 4.9 m/s^2 *.36 s^2 = 1.764 m The altitude the 1st ball will be at after 0.6 seconds is 0.6 s * 25 m/s - 1.764m = 15 m - 1.764 m The altitude the 2nd ball will be at after 0.6 seconds is 15 m - 1.764 m Both are the same value so the answer is verified.