A stone is dropped in water from a bridge 44 cm above the water. Another stone is thrown straight down 1.0 s after the first one dropped. Both strike the water at the same time. What was the initial speed of the second stone?
first stone: .44=1/2 g t^2
second stone: .44=v(t-1)+1/2 g (t-1)^1
use equation 1) TO solve for t, then put it in equation 2) and solve for v.
who was the father of physics
Galileo Galilei
To find the initial speed of the second stone, we can use the kinematic equation for vertically downward motion:
h = (1/2)gt² + vt
where h is the height (44 cm), g is the acceleration due to gravity (-9.8 m/s²), t is the time (1.0 s), and v is the initial velocity that we want to find.
First, let's convert the height from centimeters to meters: 44 cm = 0.44 m.
Substituting the values into the equation, we get:
0.44 = (1/2)(-9.8)(1.0)² + v(1.0)
Simplifying, we have:
0.44 = -4.9 + v
Rearranging the equation, we get:
v = 0.44 + 4.9
v ≈ 5.34 m/s
Therefore, the initial speed of the second stone is approximately 5.34 m/s.