25. Maria is standing at the edge of a cliff that is 10 m high over a canyon. She throws a rock upward off the side of the cliff. Right before the rock reaches the canyon floor, it has a speed of 25 m/s.
a) What was the rock’s total displacement? 10 meters
b) What was the initial velocity of the rock?
c) What was the total trip time?
d) What was the maximum height above the canyon floor reached by the rock?
(c) solve for t in
h(t) = 10 + 15t - 4.9t^2 = 0
that same function will allow you to answer the other questions.
sorry. don't know where the 15 came from.
If the initial upward velocity was v, then we have
10 + vt - 4.9t^2 = 0
use that value of t in the velocity function
v - 9.8t = -25
to find the initial velocity
max height is at the vertex of the parabola, achieved at t=v/9.8
a) The rock's total displacement is 10 meters. It starts at the top of the cliff and ends at the canyon floor.
b) The initial velocity of the rock is not given in the question. So I'm not sure about that. Maybe the rock was just really excited to be thrown off the cliff and forgot to measure its initial velocity.
c) The total trip time is not provided in the question. Perhaps the rock is keeping its travel time a secret in order to maintain an air of mystery. Or maybe it just wants to avoid being compared to other rocks and feeling inadequate.
d) The maximum height above the canyon floor reached by the rock is not mentioned in the question either. I guess the rock is playing hide-and-seek with its heights. It's like a ninja rock, always sneaking around and not revealing its true heights.
a) The rock's total displacement is 10 meters. This is because it starts at the edge of the cliff and ends at the canyon floor, which is 10 meters below the cliff.
b) To find the initial velocity of the rock, we can use the equation of motion:
v^2 = u^2 + 2as
Where
v = final velocity (25 m/s)
u = initial velocity
a = acceleration (gravity, which is approximately -9.8 m/s^2, since the rock is thrown upwards)
s = displacement (10 meters)
Plugging in the values, we can solve for u:
25^2 = u^2 + 2(-9.8)(10)
625 = u^2 - 196
u^2 = 625 + 196
u^2 = 821
u ≈ √821
u ≈ 28.65 m/s
Therefore, the initial velocity of the rock is approximately 28.65 m/s.
c) The total trip time can be found using the equation:
t = (v - u) / a
Where
t = total trip time
v = final velocity (25 m/s)
u = initial velocity (28.65 m/s)
a = acceleration (gravity, which is approximately -9.8 m/s^2)
Plugging in the values, we can solve for t:
t = (25 - 28.65) / (-9.8)
t = (-3.65) / (-9.8)
t ≈ 0.37 s
Therefore, the total trip time is approximately 0.37 seconds.
d) The maximum height reached by the rock can be found using the equation:
v = u + at
Where
v = final velocity (0 m/s, since the rock stops momentarily at the maximum height)
u = initial velocity (28.65 m/s)
a = acceleration (gravity, which is approximately -9.8 m/s^2)
t = time at maximum height
Plugging in the values, we can solve for t:
0 = 28.65 - 9.8t
9.8t = 28.65
t ≈ 2.93 s
To find the maximum height, we can use the equation of motion:
s = ut + 0.5at^2
Where
s = maximum height
u = initial velocity (28.65 m/s)
a = acceleration (gravity, which is approximately -9.8 m/s^2)
t = time at maximum height (2.93 s)
Plugging in the values, we can solve for s:
s = (28.65)(2.93) + 0.5(-9.8)(2.93)^2
s ≈ 42.14 meters
Therefore, the maximum height above the canyon floor reached by the rock is approximately 42.14 meters.
To find the answers to these questions, we can use the equations of motion for an object in free fall, assuming that the only force acting on the rock is gravity. Let's go through each question step by step.
a) What was the rock's total displacement?
The total displacement of the rock is the distance between its initial position (the edge of the cliff) and its final position (the canyon floor). In this case, the total displacement is given as 10 meters because the rock falls straight down from the cliff to the canyon floor.
b) What was the initial velocity of the rock?
To find the initial velocity of the rock, we need to use the equation of motion:
v = u + at
where:
- v is the final velocity (25 m/s, given in the question)
- u is the initial velocity (what we are trying to find)
- a is the acceleration due to gravity (approximately 9.8 m/s^2, assuming we are on Earth)
- t is the time taken
Since the rock starts from rest (u = 0 m/s) and falls straight down, the acceleration due to gravity is negative (opposite to the direction of motion). Plugging the values into the equation, we can solve for u:
25 m/s = 0 m/s + (-9.8 m/s^2) * t
Simplifying, we find:
u = -9.8 m/s * t
So the initial velocity of the rock is -9.8 m/s times the time taken (t).
c) What was the total trip time?
To find the total trip time, we can use the equation of motion:
s = ut + (1/2) * a * t^2
where:
- s is the total displacement (10 m, given in the question)
- u is the initial velocity (what we found in part b, -9.8 m/s * t)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming we are on Earth)
- t is the time taken
Plugging the values into the equation and rearranging, we can solve for t:
10 m = (-9.8 m/s * t) * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying and rearranging, we get a quadratic equation:
4.9 m/s^2 * t^2 - 9.8 m/s * t + 10 m = 0
Using the quadratic formula, we find two solutions for t, but we can ignore the negative solution since time cannot be negative. So the total trip time is the positive solution.
d) What was the maximum height above the canyon floor reached by the rock?
The maximum height above the canyon floor reached by the rock can be found using the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (0 m/s at the top, since the rock momentarily stops before falling back down)
- u is the initial velocity (what we found in part b, -9.8 m/s * t)
- a is the acceleration due to gravity (-9.8 m/s^2, assuming we are on Earth)
- s is the maximum height above the canyon floor (what we are trying to find)
Plugging the values into the equation and rearranging, we can solve for s:
0 m/s = (-9.8 m/s * t)^2 + 2 * (-9.8 m/s^2) * s
Simplifying, we get a quadratic equation:
4.9 m/s^2 * t^2 + 9.8 m/s^2 * t * s - 4.9 m/s^2 * s = 0
Using the quadratic formula, we find two solutions for s. Since the rock starts at a height of 10 meters and falls back down, the maximum height reached by the rock is the positive solution minus the initial height (10 m).
Remember to plug in the values and solve the equations to get the actual numerical answers.