- In the first case stone is thrown upwards. So first it goes up against gravity and then comes down. So its motion from going up to coming down to original throwing position is a projectile motion.
Applying equations of motion:
V= u+at
Let u= u1
a = g = 9.8m/sec2
We divide t1 into two timings , one for projectile motion and second part a stone travels from original position to ground.
So, t1=t''+t"
t'' = 2u1/g
t" = t2 as after stones reaches back original position it has velocity u1.
t1=2u1/g+t2
- In case 2 stone is thrown down with u1 velocity. Time taken is t2. Let us say distance covered by stone before reaching ground be S.
S= u1*t2+0.5*g*t2*t2 .......................................(1)
- In case there u=0
So S= 0.5*g*t3*t3................................................(2)
- Equating (1) & (2) and multiplying 2/(g*t2) on both sides we have:
Solution : t3=sqrt(t1*t2)