Question

From top of a tower, a stone is thrown up and it reaches the ground in time t1. A second stone is thrown down with the same speed and it reaches the ground in time t2. A third stone is released from rest and it reaches the ground in time t3. Then:
a) t3=1/2(t1+t2)
b)t3=sqrt(t1t2)
c)1/t3=1/t2-1/t1
d)t3^2=t1^2-t2^2

Answer 1

If v is the velocity with which the stone is thrown upward, h is the height of the tower, and g is the acceleration due to gravity, then you have three equations describing the various times.
[eqn1] 0 = h - (g t1^2)/2 + t1 v
[eqn2] 0 = h - (g t2^2)/2 - t2 v
[eqn3] 0 = h - (g t3^2)/2

Solving [eqn3] for h gives us a substitution we can use in the first two equations. These are now
.. 0 = 1/2 (g (-t1^2 + t3^2) + 2 t1 v),
.. 0 = 1/2 (g (-t2^2 + t3^2) - 2 t2 v)

Solving each of these for v and equating the results gives
.. g (t1^2 - t3^2)/(2 t1) = g (t3^2 - t2^2)/(2 t2)

This simplifies to
.. t2(t1^2 - t3^2) = t1(t3^2 - t2^2) ... multiply by 2*t1*t2/g
.. t3^2(t1+t2) = t1*t2^2 + t2*t1^2 = t1*t2*(t1+t2) ... collect t3 terms on one side
.. t3 = √(t1*t2) ... divide by (t1+t2), take the root.

Your solution is: T3 is the geometric mean of T1 and T2.