A rock is dropped from a bridge to the water below. It takes 2.4s to hit the water. Find the final speed of the rock when it hits the water. How high is the bridge above the water

See your 2:23pm post for solutin.

To find the final speed of the rock when it hits the water, we can use the equation for freefall motion:

v = u + gt

where:
v is the final velocity of the rock,
u is the initial velocity (which is 0 m/s as the rock is dropped),
g is the acceleration due to gravity (which is approximately 9.8 m/s²),
t is the time taken (which is 2.4 seconds in this case).

Substituting the values into the equation:

v = 0 + (9.8)(2.4)
v = 23.52 m/s

Therefore, the final speed of the rock when it hits the water is 23.52 m/s.

To calculate the height of the bridge above the water, we can use the equation for vertical motion:

h = ut + 1/2gt²

where:
h is the height (which is what we are trying to find),
u is the initial velocity (which is 0 m/s as the rock is dropped),
g is the acceleration due to gravity (which is approximately 9.8 m/s²),
t is the time taken (which is 2.4 seconds in this case).

Since we are trying to find the height (h), we need to solve for it. Rearranging the equation, we get:

h = 1/2gt²

Substituting the values into the equation:

h = 1/2(9.8)(2.4)²
h = 1/2(9.8)(5.76)
h = 1/2(56.448)
h = 28.224 m

Therefore, the height of the bridge above the water is 28.224 meters.