Galileo dropped a light rock and a heavy rock from the leaning tower of pisa, which is about 55 m high. suppose that galileo dropped one rock 0.50 s before the second rock. with what initial velocity should he drop the second rock so that it reaches the ground at the same time as the first rock?

To solve this problem, we make use of the equations for linear motion. The relevant formula to use here is:

y = vi t + 0.5 a t^2 - - - > 1

where,

t = time, y = distance, a = acceleration = gravity, vi = initial velocity

Let us say that the light rock is 1, and the heavy rock is 2. We know that the distance of the two rocks must be equal, therefore:

y1 = y2

vi1 t1 + 0.5 g t1^2 = vi2 t2 + 0.5 g t2^2

From the given values, we know that rock 1 is simply dropped therefore vi1 = 0, therefore:

0.5 g t1^2 = vi2 t2 + 0.5 g t2^2 - - - > 2

We also know that t1 = t2 + 0.5 or t2 = t1 - 0.5. Therefore we need first to find for the value of t1. By using equation 1:

55 = 0.5 (9.8) t1^2

t1^2 = 11.22

t1 = 3.35 s

Therefore:

t2 = t1 - 0.5 = 3.35 - 0.5

t2 = 2.85 s

Going back to equation 2:

0.5 (9.8) (3.35) ^2 = vi2 (2.85) + 0.5 (9.8) (2.85) ^2

2.85 vi2 = 15.19

vi2 = 5.33 m/s

Therefore he must throw the rock at an initial velocity of 5.33 m/s to reach the 1st rock.