a 2.50 N rock is thrown down from a cliff at 10.0 m/s.
a. what is the velocity after falling 25.0 m?
b. velocity as it hits the water 125 m below?
c. kinetic energy as it hits the H20?
To answer these questions, we can use the principles of motion and the equations of motion to find the desired values. Let's break down each question and explain how to get the answers:
a. What is the velocity after falling 25.0 m?
To find the final velocity after falling a certain distance under gravity, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
s = distance fallen
We are given:
u = 10.0 m/s (thrown downward)
s = 25.0 m (distance fallen)
Plugging in the values, we can find the final velocity (v):
v^2 = (10.0 m/s)^2 + 2 * (-9.8 m/s^2) * 25.0 m
v^2 = 100.0 m^2/s^2 - 490.0 m^2/s^2
v^2 = -390.0 m^2/s^2
Taking the square root of both sides, we get:
v = -19.7 m/s (rounded to two decimal places)
So, the velocity after falling 25.0 m is approximately -19.7 m/s (downward direction).
b. What is the velocity as it hits the water 125 m below?
To find the velocity when the rock hits the water, we can use the equation of motion given above. Since we know the initial velocity (10.0 m/s downward) and the distance fallen (125 m), we can calculate the final velocity (v):
v^2 = (10.0 m/s)^2 + 2 * (-9.8 m/s^2) * 125 m
v^2 = 100.0 m^2/s^2 - 2450.0 m^2/s^2
v^2 = -2350.0 m^2/s^2
Taking the square root:
v = -48.4 m/s (rounded to one decimal place)
So, the velocity as it hits the water 125 m below is approximately -48.4 m/s (downward direction).
c. What is the kinetic energy as it hits the water?
The kinetic energy (KE) of an object can be calculated using the formula:
KE = 0.5 * m * v^2
Where:
m = mass of the object
v = velocity of the object
However, we are only given the weight of the rock, not the mass. To calculate the mass, we can use the equation:
weight (W) = mass (m) * gravitational acceleration (g)
W = 2.50 N (given)
g = 9.8 m/s^2
Thus:
2.50 N = m * 9.8 m/s^2
Solving for m:
m = 2.50 N / 9.8 m/s^2
m ≈ 0.255 kg (rounded to three decimal places)
Now, we can substitute the values of m and v (from part b) into the kinetic energy formula:
KE = 0.5 * 0.255 kg * (-48.4 m/s)^2
KE = 0.2478 kg m^2/s^2
So, the kinetic energy as the rock hits the water is approximately 0.2478 Joules.
I am not going to do all of these for you. This and the previous one are basically the same.
The mass does not matter until we get to part c
define velocity and distance positive down
Vi = 10
g = 9.81
distance at 0 = 0
v = Vi + g t
d = Vi t + (1/2) g t^2
at 25 meters
25 = 10 t + 4.9 t^2
4.9 t^2 + 10 t -25 = 0
t = [ -10 +/- sqrt(100+ 490)]/9.81
t = [-10 +/- 24.3 ]/9.81
t = 1.46 seconds
so
v = 10 + g t = 24.3 m/s at 25 m
at water
d = 125
125 = 10 t + 4.9 t^2
4.9 t^2 + 10 t - 125 = 0
t = [ -10 +/- sqrt (100+2450)]/9.81
t = 4.13 seconds
so
v = 10 + 9.81(4.13) = 50.5 m/s at 125 m
Ke = (1/2) m v^2
= (1/2)(2.5/9.81)(50.5)^2
= 325 Joules