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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?

Respuesta :

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.

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