A 0.15-kg apple falls off a tree branch that is 2.8 m above the thick grass. The apple sinks 0.066 m into the grass while stopping. Determine the average contact force that the grass alone exerts on the apple while stopping it. Ignore air resistance.

We must solve this problem in two parts, the first with energy to find the speed with which it reaches the grass and with impulse to determine the strength in the grass

Let's look for the speed of arrival to the grass, write the energy in two points

Initial

Em₀ = U = m g h

Point 2, just by touching the grass

Em₂ = K = ½ m v²

Em₀ = Em₂

m g h = ½ m v²

v = √ 2gh

v = √( 2 9.8 2.8)

v = 7.4 m/s

Let's calculate the momentum until it stops

I = F t = m vf -m v₀

I = 0-0.15 (-7.4)

I = 1.11 N s

We must find the time during which the apple stops, let's use kinematics to find the average acceleration

[tex]v_{f}[/tex] = v = 7.4 m/s

[tex]v_{f}[/tex]² = v₀² + 2 a y

0 = v₀² + 2 a y

a = v₀² / 2y

a = 7.4²/2 0.066

a = 414.85 m / s²

Now let's look for the time to stop with this acceleration

[tex]v_{f}[/tex] = v₀ + at

0 = v₀ + at

t = v₀ / a

t = 7.4 / 414.85

t = 0.0178 s

Let's calculate with the momentum

I = F t

F = I / t

F = 1.11 / 0.0178

F = 62.4 N