A projectile of mass 1.450 kg is shot straight up with an initial speed of 20.0 m/s.

(a) How high would it go if there were no air friction?

(b) If the projectile rises to a maximum height of only 13.9 m, determine the magnitude of the average force due to air resistance.

The initial KEnergy is 1/2 m v^2

The PEnergy at to top must be equal to the initial KEnergy.
mgh=1/2 mv^2 solve for h. Notice m divides out.

b) forceresistance*13.9=1/2 m v^2
solve for the force.

a) 20.4085

b) 20.8633

bob's formula for B) is wrong...

Should be mgh-forceresistance = 1/2 m Vi^2

nah bruh

(a) To find the maximum height reached by the projectile in the absence of air friction, we can use the principles of projectile motion.

When a projectile is shot straight up, its initial vertical velocity (v₀) is positive, while its final vertical velocity (v) at the maximum height is zero. The acceleration due to gravity (g) acts in the opposite direction to the initial velocity.

The equations we will use are:

v = v₀ - gt
Δy = v₀t - (1/2)gt²

Since the maximum height is reached when v = 0, we can set the final velocity to zero and solve for the time it takes to reach that point:

0 = v₀ - gt
t = v₀ / g

Now we can substitute this value of time into the equation for the change in height:

Δy = v₀ * (v₀ / g) - (1/2)g * (v₀ / g)²
Δy = (v₀² / g) - (v₀² / 2g)
Δy = v₀² / 2g

Plugging in the given values:
v₀ = 20.0 m/s
g = 9.8 m/s²

Δy = (20.0²) / (2 * 9.8)
Δy = 400 / 19.6
Δy ≈ 20.41 m

Therefore, in the absence of air friction, the projectile would reach a height of approximately 20.41 m.

(b) To determine the magnitude of the average force due to air resistance, we can use the work-energy principle. The work done by the force of air resistance is equal to the change in the projectile's mechanical energy.

The mechanical energy of the projectile at the maximum height is given by:

E = mgh

where:
m = mass of the projectile (1.450 kg)
g = acceleration due to gravity (9.8 m/s²)
h = maximum height (13.9 m)

So, the mechanical energy is:
E = (1.450 kg) * (9.8 m/s²) * (13.9 m)
E ≈ 194.173 J

Now, we can calculate the work done by the force of air resistance. Since the work-energy principle states that work done is equal to the change in mechanical energy, we have:

Work = E - ΔKE

where ΔKE is the change in kinetic energy. At the maximum height, the projectile's velocity is zero, so ΔKE is the initial kinetic energy (KE₀):

ΔKE = KE₀ = (1/2)mv₀²

Substituting the given values:
m = 1.450 kg
v₀ = 20.0 m/s
ΔKE = (1/2) * (1.450 kg) * (20.0 m/s)²

ΔKE ≈ 290 J

Now we can calculate the work done by air resistance:

Work = E - ΔKE
Work ≈ 194.173 J - 290 J
Work ≈ -95.827 J

The negative sign indicates that the work done by air resistance is in the opposite direction to the displacement of the projectile.

The magnitude of the average force due to air resistance can be calculated using the equation:

Work = force * distance

Rearranging the equation:

force = Work / distance

In this case, the distance is the maximum height (13.9 m), so we have:

force = -95.827 J / 13.9 m
force ≈ -6.90 N

Therefore, the magnitude of the average force due to air resistance is approximately 6.90 N.