A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 35 m high. When it hits the ground at the base of the cliff the rock has a speed of 28 m/s.
(a) Assuming that air resistance can be ignored, find the initial speed of the rock.
(b) Find the greatest height of the rock as measured from the base of the cliff.
(a) When air resistance can be ignored, the sum of potential and kinetic energy is constant. That means that V^2/2 + g h is constant, where h is the height measured from any reference plane. (The mass M cancels out). Use that fact to solve for the initial velocity Vo.
28^2/2 = Vo^2/2 + g*35
(b) At the greatest height, V = 0. use the same approach to solve for Hmax
To solve this problem, we can use the principles of projectile motion.
(a) To find the initial speed of the rock, we can use the conservation of energy. The potential energy at the top of the cliff is equal to the kinetic energy when it hits the ground.
Let's calculate the potential energy (PE) at the top of the cliff and the kinetic energy (KE) when it hits the ground.
PE = m * g * h
where m is the mass of the rock (0.26 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the cliff (35 m).
PE = 0.26 * 9.8 * 35
PE = 89.47 Joules
KE = 1/2 * m * v^2
where v is the final velocity of the rock (28 m/s).
KE = 0.5 * 0.26 * 28^2
KE = 101.92 Joules
Since the energy is conserved, the initial potential energy is equal to the final kinetic energy.
PE = KE
89.47 = 101.92
101.92 = 89.47
KE = 89.47 Joules
Now we can calculate the initial speed (vi) using the equation for kinetic energy:
KE = 1/2 * m * vi^2
89.47 = 0.5 * 0.26 * vi^2
Solving for vi:
vi^2 = (2 * 89.47) / 0.26
vi^2 = 689.46
vi = √689.46
vi ≈ 26.24 m/s
Therefore, the initial speed of the rock is approximately 26.24 m/s.
(b) To find the greatest height of the rock as measured from the base of the cliff, we can use the equations of motion for vertical motion.
The final velocity at the top of the cliff (when the rock reaches its maximum height) is 0 m/s.
Using the equation:
vf^2 = vi^2 + 2 * g * Δy
where vf is the final velocity (0 m/s), vi is the initial velocity (26.24 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and Δy is the change in height (the greatest height we want to find).
0 = 26.24^2 + 2 * (-9.8) * Δy
Simplifying the equation:
0 = 688.3872 - 19.6 * Δy
Rearranging the terms:
19.6 * Δy = 688.3872
Solving for Δy:
Δy = 688.3872 / 19.6
Δy ≈ 35.11 m
Therefore, the greatest height of the rock as measured from the base of the cliff is approximately 35.11 m.
To answer these questions, we can use the equations of motion for objects in free fall. Let's start by finding the initial speed of the rock.
(a) The initial speed of the rock can be found using the equation:
v² = u² + 2as,
where:
v is the final velocity (28 m/s),
u is the initial velocity (what we want to find),
a is acceleration (which can be taken as -9.8 m/s², since the rock is thrown upwards),
and s is displacement (35 m).
Plugging in these values, the equation becomes:
(28 m/s)² = u² + 2(-9.8 m/s²)(35 m).
This simplifies to:
784 m²/s² = u² - 686 m²/s².
Rearranging and solving for u:
u² = 784 m²/s² + 686 m²/s²,
u² = 1470 m²/s²,
u = √1470 m/s.
Therefore, the initial speed of the rock is approximately 38.3 m/s.
(b) To find the greatest height of the rock, we can use the equation:
v² = u² + 2as,
where:
v is the final velocity (0 m/s when the rock reaches its maximum height),
u is the initial velocity (38.3 m/s, which we found in part (a)),
a is acceleration (-9.8 m/s²),
and s is the displacement (what we want to find).
Plugging in these values, the equation becomes:
(0 m/s)² = (38.3 m/s)² + 2(-9.8 m/s²)s.
This simplifies to:
0 m²/s² = 1468.89 m²/s² - 19.6 m/s² * s.
Rearranging and solving for s:
19.6 m/s² * s = 1468.89 m²/s²,
s = 1468.89 m²/s² / 19.6 m/s²,
s = 75 m.
Therefore, the greatest height of the rock, measured from the base of the cliff, is approximately 75 meters.