Question

A stone is thrown up vertically from the ground (the gravitational acceleration is g = 10 m/s^2 ). After a time dt =1 s, a second stone is thrown up vertically. The first stone has an initial speed v1 = 11.0 m/s, and the second stone v2 = 16.0 m/s.

(a) At what time after the first stone is thrown will the two stones be at the same altitude above ground? (in seconds)

Answer 1

a. h1 = V1*t + 0.5g*t^2
h1 = 11*1 - 5*1^2 = 6 m. Head start.
V^2 = V1^2 + 2g*h
V^2 = 11^2 - 20*6 = 1
V = 1.0 m/s.

h2=V2*t + 0.5g*t^2 = 6 + V*t + 0.5g*t^2
16*t - 5*t^2 = 6 + 1*t - 0.5g*t^2
The 0.5g*t^2 terms cancel:
16t - t = 6
15t = 6.0
t = 0.4 s. to catch up.