Question

A stone is projected vertically upwards with a velocity of 20m/s. 2s later another stone is similarly projecte?
with the same velocity. when the two stones meet the second one is rising at a velocity of 10m/s.neglecting air resistance calculate the length of time the second stone is in motion before they met.
the velocity of the first stone when they met.take gravity to be 10m/s.

Answer 1

Stone 1
h = 0 + 20 t - 5 t^2
v = 20 - 10 t

Stone 2
h = 0 + 20(t-2) - 5 (t^2 - 4 t + 4)
v = 20 - 10(t-2)
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at h, v of stone 2 = +10
10 = 20 -10 t + 20
10 t = 30
t = 3 seconds that the first one is in the air
t-2 = 1 second for the second one (first question)
v = 20 - 10*3 = -10 m/s for the first stone at collision

Answer 2

Stone #1:
V = Vo + g*t = 20 + (-10)*2 = 0 m/s.(h = h max.).

Stone #2:
a. V = Vo + g*t = 10.
20 + (-10)t = 10,
t = 1 s.

Stone #1:
b. V = Vo + g*t = 0 + 10*(3-2) = 10 m/s. = Velocity of 1st. stone when they
met.
Note: The 1st. stone had been in air 3 s(1+2) when they met. But 2 of those
3 s was rise time, That left 1 s fall time.

Answer 3

Thanks but i need more explanation

Answer 4

What is the answer on the first question?

Answer 5

Suggest assumptions you made in calculating