A projectile is fired at an upward angle of 55.0 degrees from the top of a 230m cliff with a speed of 180m/s.What will be its speed when it strikes the ground below? (Use conservation of energy.)

To determine the speed of the projectile when it strikes the ground below, we can use the conservation of energy principle. The conservation of energy states that the initial mechanical energy of the system is equal to the final mechanical energy of the system, assuming there are no external forces acting on the projectile.

In this case, the initial mechanical energy of the projectile is equal to its gravitational potential energy at the top of the cliff (230m above the ground) and its kinetic energy when it is fired. The final mechanical energy of the projectile is equal to its gravitational potential energy just before hitting the ground and its kinetic energy at that point.

First, we need to find the initial and final gravitational potential energy of the projectile. The gravitational potential energy is given by the formula: PE = m * g * h, where m is the mass of the projectile, g is the acceleration due to gravity, and h is the height.

Given:
Height of the cliff (h) = 230m
Angle of projection = 55.0 degrees
Speed of projection = 180m/s

We can find the initial and final gravitational potential energy as follows:

Initial Gravitational Potential Energy:
PE_initial = m * g * h
PE_initial = m * 9.8 m/s^2 * 230 m

Final Gravitational Potential Energy:
PE_final = m * g * h
PE_final = m * 9.8 m/s^2 * 0 m (since the projectile has hit the ground)

Now, let's find the initial and final kinetic energy of the projectile.

Initial Kinetic Energy:
KE_initial = 0.5 * m * v^2 (where v is the speed of projection)
KE_initial = 0.5 * m * (180 m/s)^2

Final Kinetic Energy:
KE_final = 0.5 * m * v_final^2 (where v_final is the speed at which the projectile strikes the ground)

According to the conservation of energy principle:
PE_initial + KE_initial = PE_final + KE_final

Since the projectile starts at rest vertically (in the y-direction), the initial kinetic energy (KE_initial) is 0.

Thus, the conservation of energy equation becomes:
PE_initial = PE_final + KE_final

This simplifies to:
m * g * h = m * g * 0 + 0.5 * m * v_final^2

Simplifying further:
g * h = 0.5 * v_final^2

Now we can solve for v_final:

v_final^2 = (2 * g * h)
v_final = sqrt(2 * g * h)

Substituting the known values:
v_final = sqrt(2 * 9.8 m/s^2 * 230 m)

Calculating the value of v_final:
v_final = sqrt(4504)
v_final ≈ 67.1 m/s

Therefore, when the projectile strikes the ground below, its speed will be approximately 67.1 m/s.

You won't need to use the angle information.

What don't you understand about using conservation of energy?

Set the loss of potential energy (M g H) equal to the INCREASE in kinetic energy (1/2) M V^2. The mass will cancel out, so you don't need to know it.

Vf^2 - Vi^2 = 2 g H

H = 230 m and Vi = 180 m/s

Compute the final velocity Vf