A particle, starting from point A in the drawing at a height h0 = 2.3 m, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height hf = 4.6 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

m/s

To find the speed of the particle at point A, we can make use of the conservation of energy principle. We can equate the initial potential energy at point A to the final kinetic energy at point A.

The potential energy at point A is given by:
Potential energy (PE) = mass of the particle (m) * acceleration due to gravity (g) * height at point A (h0)

The kinetic energy at point A is given by:
Kinetic energy (KE) = 1/2 * mass of the particle (m) * speed at point A (v)^2

Since there is no friction or air resistance, the total mechanical energy is conserved. Therefore, we can equate the potential energy at point A to the sum of the kinetic energy at point A and the potential energy at point B.

PE at A = KE at A + PE at B

Plugging in the given values:
m * g * h0 = 1/2 * m * v^2 + m * g * hf

Simplifying the equation:
v^2 = 2 * g * (hf - h0)

Taking the square root of both sides:
v = √(2 * g * (hf - h0))

Plugging in the values for the acceleration due to gravity (g = 9.8 m/s^2), the initial height (h0 = 2.3 m), and the final height (hf = 4.6 m):

v = √(2 * 9.8 * (4.6 - 2.3)) = √(2 * 9.8 * 2.3) = √(45.24) = 6.72 m/s

Therefore, the speed of the particle at point A is 6.72 m/s.

To find the speed of the particle at point A, we can use the principle of conservation of energy. Since we are ignoring friction and air resistance, the total mechanical energy of the particle remains constant throughout its motion.

The total mechanical energy at point A is the sum of the particle's kinetic energy (KE) and potential energy (PE) due to its height.

The kinetic energy at point A is given by KE = 1/2 * mv^2, where m is the mass of the particle and v is its velocity.

The potential energy at point A is given by PE = mgh, where m is the mass of the particle, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the particle above the reference point (floor).

Since the particle is projected downward, its initial velocity at point A is negative. Therefore, we can rewrite the kinetic energy as KE = -1/2 * mv^2.

At point B, the particle reaches its maximum height hf = 4.6 m. At this height, the particle's velocity is momentarily zero. Therefore, the potential energy at point B is given by PE = mghf.

Since the total mechanical energy remains constant, we can set the sum of the kinetic energy and potential energy at points A and B equal to each other:

-1/2 * mv^2 + mgh0 = mghf

Simplifying and rearranging the equation, we get:

-1/2 * v^2 + gh0 = ghf

Now we can plug in the known values:
g = 9.8 m/s^2
h0 = 2.3 m
hf = 4.6 m

Since we want to find the speed at point A, we are looking for the magnitude of the velocity v. Therefore, we can remove the negative sign and solve for v^2:

1/2 * v^2 = ghf - gh0

Next, we can multiply both sides of the equation by 2 to eliminate the fraction:

v^2 = 2ghf - 2gh0

Finally, taking the square root of both sides of the equation, we can solve for v:

v = sqrt(2ghf - 2gh0)

By substituting the values of g, hf, and h0 into this equation, we can calculate the speed of the particle at point A in m/s.