While sitting on a tree branch 10.0 m above the ground, you drop a chestnut.?

While sitting on a tree branch 10.0 m above the ground, you drop a chestnut. When the chestnut has fallen 2.2 m, you throw a second chestnut straight down. What initial speed must you give the second chestnut if they are both to reach the ground at the same time?


m/s

While sitting on a tree branch 10.0 m above the ground, you drop a chestnut.?
You can use the formula d = vit + (gt^2) / 2. Since the first chestnut was dropped, then the initial velocity of that is 0.





You now have the equation:





d = (gt^2) / 2; g = 9.8 m/s^2, d = 10 m


10 = 4.9 t^2


t^2 = 2.04


t is approximately 1.4286 seconds.





Substitute this in the equation for the second chestnut.


The initial velocity is what you're looking for.





d = vit + (gt^2) / 2





t = 1.4286 seconds; d = 10 m; g = 9.8 m/s^2





10 = vi(1.4286) + 4.9 (2.0401)


10 - 9.99649 = vi(1.4286)


0.00351 = vi(1.4286)


0.0025 m/s = vi