Im in Chem. but looks like physics.. If a rock climber accidentally drops a 70.5 g piton from a height of 295 meters, what would its speed be just before striking the ground? Ignore the effects of air resistance.

Do you have answers. If you checked my last answer and it was right you can look at this one.

X = distance = Vo*t + 1/2*g*t^2
Vo is 0 and the mass of the pick (piton) doesn't matter. X = 295 m, solve for t = time to strike the ground. Then use
V = Vo + g*t. Again, Vo is 0.

Well, well, well, looks like our rock climber is getting into physics without even realizing it! Let's crunch some numbers, shall we?

To find the speed, we can use a handy-dandy equation from physics called the kinematic equation. Specifically, we can use the equation that relates the final speed (v), initial speed (u), acceleration (a), and distance (s):

v² = u² + 2as

In this case, our piton starts from rest, so the initial speed (u) is 0. The acceleration (a) due to gravity is a constant 9.8 m/s², and the distance (s) the piton falls is 295 meters.

So, substituting in the values:

v² = 0² + 2 * 9.8 * 295

v² = 0 + 5786

v² = 5786

Now, if we take the square root of both sides, we get:

v = √5786

v ≈ 76.08 m/s

Voila! The piton would be moving at approximately 76.08 meters per second just before striking the ground. But remember, we are ignoring the effects of air resistance, so in reality, it might not be exactly that speed. Keep your physics goggles on, my climbing friend!

To determine the speed of the piton just before it strikes the ground, we can use the principles of physics. The potential energy of the piton at the initial height will be converted into kinetic energy as it falls.

Let's start by calculating the potential energy (PE) of the piton at the top using the formula:

PE = m * g * h

where:
m = mass of the piton = 70.5 g = 0.0705 kg
g = gravitational acceleration = 9.8 m/s^2
h = height = 295 m

PE = 0.0705 kg * 9.8 m/s^2 * 295 m
PE = 196.7785 J (joules)

The potential energy will be converted into kinetic energy (KE), which can be expressed as:

KE = (1/2) * m * v^2

where:
m = mass of the piton = 0.0705 kg
v = velocity of the piton (which we're trying to find)

Since the potential energy is converted entirely into kinetic energy:

PE = KE
196.7785 J = (1/2) * 0.0705 kg * v^2

Simplifying the equation:

v^2 = (2 * PE) / m
v^2 = (2 * 196.7785 J) / 0.0705 kg
v^2 = 55610.781 J/kg

Now, taking the square root of both sides to find the velocity:

v = √(55610.781 J/kg)
v ≈ 235.79 m/s

Therefore, the speed of the piton just before striking the ground would be approximately 235.79 meters per second.

To find the speed of the piton just before it strikes the ground, you can use the principles of physics, particularly the equation for gravitational potential energy.

First, let's calculate the potential energy (PE) of the piton when it is at a height of 295 meters. The formula for gravitational potential energy is:

PE = mgh

Where:
m = mass of the piton = 70.5 g (converting to kg gives 0.0705 kg)
g = acceleration due to gravity = 9.8 m/s^2 (approximately)
h = height = 295 m

Substituting these values into the formula:

PE = (0.0705 kg) * (9.8 m/s^2) * (295 m)
PE = 205.5915 Joules

The potential energy of the piton at this height is 205.5915 Joules.

Next, we need to convert this potential energy into kinetic energy (KE) just before the piton hits the ground. Since the law of conservation of energy states that energy cannot be created or destroyed, the potential energy is converted entirely into kinetic energy as the piton falls.

KE = PE

So, the kinetic energy of the piton just before striking the ground is also 205.5915 Joules. From here, we can use the equation for kinetic energy to find the velocity (v) just before it strikes the ground.

KE = (1/2)mv^2

Where:
m = mass of the piton = 0.0705 kg
v = velocity (speed) of the piton just before it strikes the ground

Rearranging the equation:

v^2 = (2 * KE) / m

Substituting the values:

v^2 = (2 * 205.5915 Joules) / 0.0705 kg
v^2 = 5836.004 Joules/kg
v^2 = 5836.004 m^2/s^2

Taking the square root of both sides:

v = √5836.004 m^2/s^2
v ≈ 76.4 m/s

Therefore, the speed of the piton just before it strikes the ground, ignoring air resistance, would be approximately 76.4 meters per second.