A rock is dropped of the side of a bridge and hits the water below 4s later.
a. What was the rock’s velocity when it hit the water?
b. What was the rock’s average velocity as it fell?
c. What is the height of the bridge above the water?
To solve these questions, we need to use the equations of motion for objects in free fall:
1. Find the rock's velocity when it hits the water (part a):
The initial velocity, u, of the rock is zero because it was dropped. The time, t, it takes for the rock to hit the water is given as 4s.
We can use the equation: v = u + gt, where g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Plugging in the values, we have v = 0 + (9.8 m/s^2)(4s) = 39.2 m/s.
Therefore, the rock's velocity when it hits the water is 39.2 m/s.
2. Find the rock's average velocity as it fell (part b):
The average velocity, v_avg, can be calculated using the equation: v_avg = (u + v) / 2.
Since the initial velocity, u, is zero, we can substitute v into the equation.
Thus, v_avg = (0 + 39.2 m/s) / 2 = 19.6 m/s.
Therefore, the rock's average velocity as it fell is 19.6 m/s.
3. Find the height of the bridge above the water (part c):
We can use the equation for distance fallen in free fall: d = ut + 0.5gt^2.
We know the time it takes for the rock to hit the water, t, is 4s, and the initial velocity, u, is zero.
Plugging in the values, we have d = 0(4s) + 0.5(9.8 m/s^2)(4s)^2 = 78.4 m.
Therefore, the height of the bridge above the water is 78.4 m.
To answer these questions, we can use the equations of motion to analyze the scenario. The key equation we will use is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
To find the answers to each question, we need to gather more information. Let's start with the information given:
Given:
Time taken, t = 4 seconds
To solve the questions, we need to find the values of velocity (v), initial velocity (u), acceleration (a), and height (h).
Let's solve each question step by step:
a. What was the rock’s velocity when it hit the water?
To find the final velocity (v), we need to know the initial velocity (u), which is not given. However, we can assume that the initial velocity is zero since the rock was dropped.
Given:
Initial velocity, u = 0 m/s (assuming the rock was dropped)
Next, we need the acceleration (a), which can be calculated using the equation:
a = (v - u) / t
We know the final velocity when the rock hits the water is zero since it stops moving at that point:
Final velocity, v = 0 m/s
Plugging in the values, we have:
0 = (0 - 0) / 4
0 = 0 / 4
Therefore, the rock's velocity when it hits the water is 0 m/s.
b. What was the rock’s average velocity as it fell?
Average velocity is calculated as the total displacement divided by the total time taken. In this scenario, the displacement is the height of the bridge, which is yet to be determined.
Given:
Time, t = 4 seconds
To calculate the average velocity, we need to know the displacement (distance), which is the height of the bridge above the water.
c. What is the height of the bridge above the water?
To calculate the height of the bridge above the water, we can use the equation of motion:
h = 0.5 * g * t^2
Where:
h = height of the bridge
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken (4 seconds)
Plugging in the values, we have:
h = 0.5 * 9.8 * 4^2
h = 0.5 * 9.8 * 16
h = 78.4 meters
Therefore, the height of the bridge above the water is 78.4 meters.
Now that we know the height of the bridge, we can calculate the average velocity.
Average velocity = Total displacement / Total time taken
Given:
Displacement = Height of the bridge = 78.4 meters
Time, t = 4 seconds
Average velocity = 78.4 / 4
Average velocity = 19.6 meters per second
Therefore, the rock's average velocity as it fell is 19.6 m/s.
To do this problem you just need to use a few simple equations:
V = V0 + at
X = X0 + V0*t + .5*a*(t^2)
and
V^2 = V0^2 + 2*a*(X - X0)
where V0 equals the initial velocity
X0 equals the initial height
t equals the time
a is just the acceleration (gravity in this case [9.81 m/s^2])
V is the final velocity
X is the final height
I think this should help you.
c. use kinematics equation
y=vi*t+1/2at^2
vi= 0 (because rock was dropped)
a=-9.8m/s^2
t= 4s (time given)
plug in and you get your height
a. use kinematics equation
vf^2=vi^2+2ay
vi=0 (rock dropped)
a=-9.8
y= (whatever you got from part c)
plug in and solve
b. ?