A new roller coaster is a conservative system. If the coaster leaves the starting point with a speed of 5 m/s and a height of 25 m, what is its velocity when it reaches a height of 2.5 m?

A. 9.81 m/s^2 B. 21.6 m/s C. 15.4 m/s D. 12.6 m/s

Pls show work thank u!

V^2 = Vo^2 + 2g*d.

V^2 = 25 + 19.6*(25-2.5) = 466
V = 21.6 m/s.

To find the velocity of the roller coaster when it reaches a height of 2.5 m, we can use the principle of conservation of energy.

First, we need to determine the potential energy (PE) and kinetic energy (KE) at the starting point and the height of 2.5 m.

At the starting point:
Potential Energy (PE1) = mgh1,
where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s^2), and h1 is the height (25 m).
PE1 = m * 9.81 * 25

Kinetic Energy (KE1) = 0.5 * m * v1^2,
where v1 is the initial velocity (5 m/s).
KE1 = 0.5 * m * 5^2

At a height of 2.5 m:
Potential Energy (PE2) = mgh2,
where h2 is the height (2.5 m).
PE2 = m * 9.81 * 2.5

Kinetic Energy (KE2) = 0.5 * m * v2^2,
where v2 is the velocity at the height of 2.5 m.

According to the principle of conservation of energy, the total mechanical energy (potential energy + kinetic energy) is conserved. Therefore:
PE1 + KE1 = PE2 + KE2

Substituting the formulas for potential and kinetic energy, we can solve for v2:
m * 9.81 * 25 + 0.5 * m * 5^2 = m * 9.81 * 2.5 + 0.5 * m * v2^2

Now, we can simplify the equation and solve for v2:
245.25 + 0.5 * 25 = 24.525 + 0.5 * v2^2
245.25 + 12.5 = 24.525 + 0.5 * v2^2
257.75 = 24.525 + 0.5 * v2^2
0.5 * v2^2 = 233.225
v2^2 = 233.225 / 0.5
v2^2 = 466.45
v2 = sqrt(466.45)
v2 ≈ 21.61 m/s

Therefore, the velocity of the roller coaster when it reaches a height of 2.5 m is approximately 21.61 m/s.

The correct answer is B. 21.6 m/s.