A stone is dropped from a height of 49m and simultaneously another ball is thrown upward from the ground with a speed of 40m/s.

Question

3 years ago

5

when and where do the two stone meet

1 answer:

Lilit [14] · Answer 1 · 2022-06-30T08:36:39+03:00

Answer:

S1 = 1/2 g t^2 distance stone falls in time t

S2 = Vy t - 1/2 g t^2 distance thrown stone rises in time t

H = 49 = S1 + S2 = Vy t

t = 49 / 40 sec time when stones meet

Check:

Stone 1 falls: 1/2 g t^2 = 1/2 * 9.8 * (49 / 40)^2 = 7.35 m

Stone 2 rises : 40 * (49 / 40) - 1/2 * 9.8 (49 / 40)^2 = 41.65 m