I really need help with these questions. My teacher doesn't provide me with anything to go on only some formulas and maybe 3 examples. It would be greatly appreciated!

1. Find the speed of a rock which is thrown off the top of a 20 m tall building at 15 m/s when it makes contact with a bird which is flying at an altitude of 10 m above the ground.

2. A 50 kg roller coaster car is launched, from ground level, at 20 m/s. How fast will it be moving when it reaches the top of a loop, which is 12 m above the ground?

3. A 200 g model rocket is observed to rise 100 m above the ground after launch. What must have been the launch speed of the rocket at the ground?

1.

initial kinetic energy = (1/2)m (225)
went down from 20 meters to 10 meters
potential energy went down by m g h
= m (9.81)(10)
so gain ke of m (9.81)(10)
so now total ke = (225/2) m + 98.1 m
= (1/2) m v^2
so
v^2 = 2 (113+98.1)

2. same as 1 above
ke at bottom = (1/2) m (400)
loses m (9.81)(12)
so
(1/2) v^2 = 200 + 9.81*12

3. LOL, same problem once again, your turn (must have gone straight up)

I'd be happy to help you with these questions!

1. To find the speed of the rock, you can use the equation for kinetic energy (KE). Given that the rock is thrown off the top of a 20 m tall building and makes contact with a bird flying at an altitude of 10 m above the ground, we can assume that the total vertical distance traveled by the rock is 20 m + 10 m = 30 m.

Since the bird is flying at an altitude of 10 m above the ground, we can conclude that the rock would have to fall 30 m in order to reach the same altitude as the bird.

Using the formula for potential energy (PE) considering the height of the rock and its mass (m), we have:

PE = mgh

Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

The potential energy gained by the rock will be equal to the kinetic energy it possesses just before it makes contact with the bird, assuming no energy loss during the fall. Therefore:

KE = mgh

Solving for velocity (v), we get:

KE = (1/2)mv^2
mgh = (1/2)mv^2

We can cancel out the mass (m) on both sides of the equation and solve for velocity (v):

gh = (1/2)v^2
2gh = v^2
v = sqrt(2gh)

Substituting the given values, with g = 9.8 m/s^2 and h = 30 m, we can calculate the speed of the rock.

2. To determine how fast the roller coaster car will be moving when it reaches the top of the loop, we can apply the conservation of energy principle. Before the car reaches the top of the loop, it has kinetic energy (KE) due to its initial velocity and potential energy (PE) due to its height above the ground.

Using the equation for potential energy (PE) and the equation for kinetic energy (KE), we have:

PE = mgh
KE = (1/2)mv^2

At the top of the loop, the car will momentarily lose contact with the track, and its velocity will become zero. This occurs because the normal force exerted by the track goes to zero, resulting in a net force equal to the gravitational force.

So, at the top of the loop, the car will have zero kinetic energy:

KE = 0

The gravitational potential energy at the top of the loop can be determined by considering the height of the loop above the ground. So:

PE = mgh

Equating the total mechanical energy (KE + PE) before and at the top of the loop, we get:

(mgh) + (1/2)mv^2 = 0

Simplifying the equation, we have:

(mgh) = -(1/2)mv^2

Canceling out the mass (m) on both sides of the equation and solving for velocity (v), we have:

gh = -(1/2)v^2
-2gh = v^2
v = sqrt(-2gh)

Substituting the given values, with g = 9.8 m/s^2 and h = 12 m, we can calculate the velocity of the roller coaster car at the top of the loop.

3. To find the launch speed of the rocket at the ground, we can use the principle of conservation of energy. At the launch site, the rocket has kinetic energy (KE) due to its initial speed and potential energy (PE) due to its height above the ground.

Using the equation for potential energy (PE) and the equation for kinetic energy (KE), we have:

PE = mgh
KE = (1/2)mv^2

At the highest point of the rocket's trajectory, its velocity becomes zero and its potential energy is at its maximum. So, at the highest point:

PE = mgh

The total mechanical energy (KE + PE) at the launch site and the highest point should be equal since there is no energy loss:

KE + PE = PE

Substituting the respective equations for KE and PE, we have:

(1/2)mv^2 + mgh = mgh

Canceling out the mass (m) on both sides of the equation and simplifying, we get:

(1/2)v^2 = 0

Solving for velocity (v), we find:

v = 0

Therefore, the launch speed of the rocket at the ground is 0 m/s.

Sure, I'd be happy to help you with these questions step by step!

1. To find the speed of the rock when it makes contact with the bird, we can use the equation for vertical motion. The initial vertical velocity is given as 15 m/s, and the rock falls from a height of 20 m to a height of 10 m. We can use the equation:

vf^2 = vi^2 + 2ad

Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance fallen. In this case, the acceleration is due to gravity, which is approximately 9.8 m/s^2. We can rearrange the equation to solve for vf:

vf = sqrt(vi^2 + 2ad)

Plugging in the values:

vf = sqrt(15^2 + 2 * 9.8 * (20 - 10))

vf = sqrt(225 + 196)

vf = sqrt(421)

vf ≈ 20.5 m/s

So the speed of the rock when it makes contact with the bird is approximately 20.5 m/s.

2. To find the speed of the roller coaster car when it reaches the top of the loop, we can again use the equation for vertical motion. The initial velocity is 20 m/s and the roller coaster car rises from ground level to a height of 12 m. We can use the same equation as before:

vf^2 = vi^2 + 2ad

Rearranging the equation to solve for vf:

vf = sqrt(vi^2 + 2ad)

Plugging in the values:

vf = sqrt(20^2 + 2 * 9.8 * 12)

vf = sqrt(400 + 235.2)

vf = sqrt(635.2)

vf ≈ 25.2 m/s

So the speed of the roller coaster car when it reaches the top of the loop is approximately 25.2 m/s.

3. To find the launch speed of the rocket on the ground, we can use the equation for vertical motion. The rocket rises to a height of 100 m and has a mass of 200 g, which is equivalent to 0.2 kg. Again, we can use the same equation:

vf^2 = vi^2 + 2ad

Rearranging the equation to solve for vi:

vi = sqrt(vf^2 - 2ad)

Since the rocket is launched vertically upward, its final velocity at the top is 0 m/s. Plugging in the values:

vi = sqrt(0 - 2 * 9.8 * 100)

vi = sqrt(-1960)

Since the square root of a negative number is undefined in this context, it means that the rocket does not have enough initial velocity to reach a height of 100 m. Therefore, the launch speed of the rocket at the ground must be greater than the calculated value.

I hope these explanations helped! Let me know if you have any further questions.