A rock is dropped from bridge to the water below. It take 2.4s to hit water fid the final speed of the rock when it hits the water. How high is the bridge above water
1. Vf = Vo + gt,
Vf = 0 + 9.8*2.4 = 23.52 m/s.
2. h = Vo*t + 0.5g*t^2,
h = 0 + 4.9*(2.4)^2 = 28.22 m.
To find the final speed of the rock when it hits the water, we can use the kinematic equation:
vf = vi + gt
Where:
vf = final velocity (which is the speed when it hits the water)
vi = initial velocity (which is 0 m/s when the rock is dropped)
g = acceleration due to gravity (approximately 9.8 m/s^2 for the Earth)
t = time taken (given as 2.4 seconds)
Plugging in the values into the equation:
vf = 0 + 9.8 * 2.4
= 23.52 m/s
So, the final speed of the rock when it hits the water is 23.52 m/s.
To find the height of the bridge above the water, we can use another kinematic equation:
d = vi * t + (1/2) * g * t^2
Where:
d = distance (which is the height of the bridge above the water)
vi = initial velocity (which is 0 m/s when the rock is dropped)
g = acceleration due to gravity (approximately 9.8 m/s^2 for the Earth)
t = time taken (given as 2.4 seconds)
Plugging in the values into the equation:
d = 0 * 2.4 + (1/2) * 9.8 * (2.4)^2
= 0 + 11.76
= 11.76 m
Therefore, the height of the bridge above the water is 11.76 meters.
To find the final speed of the rock when it hits the water, we can use the equation of motion:
v = u + gt
Where:
v = final velocity (speed) of the rock
u = initial velocity (speed) of the rock (which is 0 since it was dropped)
g = acceleration due to gravity (-9.8 m/s^2 on Earth)
t = time taken to hit the water (2.4 seconds in this case)
Plugging in the values, we have:
v = 0 + (-9.8 m/s^2) * 2.4s
v = -23.52 m/s
Since speed cannot be negative, we take the magnitude of the answer:
|v| = 23.52 m/s
Therefore, the final speed of the rock when it hits the water is 23.52 m/s.
To determine the height of the bridge above the water, we need to use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height of the bridge above the water
u = initial velocity (speed) of the rock (which is 0 since it was dropped)
g = acceleration due to gravity (-9.8 m/s^2 on Earth)
t = time taken to hit the water (2.4 seconds in this case)
Plugging in the values, we have:
h = 0 * 2.4s + (1/2)(-9.8 m/s^2) * (2.4s)^2
h = -11.76 m
Since height cannot be negative, we take the magnitude of the answer:
|h| = 11.76 m
Therefore, the height of the bridge above the water is 11.76 meters.