A rock is thrown down with an initial speed of 30.0ft/s from a bridge to the water below. It takes 3.50s for the rock to hit the water. (a) Find the speed (in ft. /s) of the rock as it hits the water. (b) How high is the bridge above the water?
To find the speed of the rock as it hits the water, we can use the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the rock is thrown down, so the initial velocity (u) is 30.0 ft/s, the acceleration (a) is the acceleration due to gravity, which is approximately 32.2 ft/s² (assuming we are neglecting air resistance), and the time (t) is 3.50 s.
(a) Find the speed (in ft/s) of the rock as it hits the water:
v = u + at
v = 30.0 + (32.2)(3.50)
v ≈ 30.0 + 112.7
v ≈ 142.7 ft/s
Therefore, the speed of the rock as it hits the water is approximately 142.7 ft/s.
To find the height of the bridge above the water, we can use the equation of motion:
s = ut + (1/2)at²
where:
s = displacement (height of the bridge)
u = initial velocity (30.0 ft/s)
a = acceleration due to gravity (32.2 ft/s²)
t = time (3.50 s)
Since the rock is thrown down, the initial velocity is positive, the acceleration due to gravity is negative, and the displacement is the height of the bridge above the water.
(b) How high is the bridge above the water:
s = ut + (1/2)at²
s = (30.0)(3.50) + (1/2)(32.2)(3.50)²
s ≈ 105.0 + (1/2)(32.2)(12.25)
s ≈ 105.0 + (1/2)(396.35)
s ≈ 105.0 + 198.18
s ≈ 303.18 ft
Therefore, the height of the bridge above the water is approximately 303.18 ft.
To find the speed of the rock as it hits the water, we can use the equation of motion for vertical free fall:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration (due to gravity) = 32.2 ft/s^2 (assuming standard gravity)
t = time = 3.50s
(a) Substituting the given values into the equation, we have:
v = 30.0 ft/s + (32.2 ft/s^2)(3.50 s)
v = 30.0 ft/s + 112.7 ft/s
v = 142.7 ft/s
Therefore, the speed of the rock as it hits the water is 142.7 ft/s.
(b) To find the height of the bridge above the water, we can use another equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (height of the bridge)
u = initial velocity
a = acceleration (due to gravity)
t = time
Since the rock is thrown downward, its initial velocity (u) is negative (-30.0 ft/s). Substituting the given values into the equation, we have:
s = (-30.0 ft/s)(3.50 s) + (1/2)(32.2 ft/s^2)(3.50 s)^2
s = -105.0 ft + (1/2)(32.2 ft/s^2)(12.25 s^2)
s = -105.0 ft + 198.0 ft
s = 93.0 ft
Therefore, the height of the bridge above the water is 93.0 ft.
a. V = Vo + g*t = 30 + 32*3.5 = 135 Ft/s
b. h = Vo*t + 0.5g*t^2
h = 30*3.5 + 16*3.5^2 = 301 Ft.