Question

A girl of mass 50kg jumps in a cirus from a height of 100m, on a net below, as a result the net sags down by 2m, find the force given by the net on the man, to prevent any injury? ​

Answer 1

Given:-

• Mass of Girl (m) = 50kg

• Height (h) = 100m

• Sags down by the girl when she jumped over the net = 2m

• Initial velocity in 1st case(u) = 0m/s

• Final Velocity in 2nd case (v) = 0m/s

To Find:-

• Force given by the net on the man (F).

Solution:-

As we know that the acceleration due to gravity is 10m/s²

Using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ v² = 0² + 2×10×100

→ v² = 0 + 2000

→ v² = 2000

→ v =√2000

→ v = 44.72m/s

Now the initial velocity of girl when she jumped on the net and the net sags down by 2m .here the Final velocity of the girl is 0m/s.

Again using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = u² + 2×10×100

→ u = √2000

here it is found that the initial velocity= Final velocity at the point where the sags down

→ u = 44.72m/s

Now Calculating the acceleration of the girl when the sags down by 2m

→ v² = u² + 2as

Put the value we get

→ 0² = (44.72)² +2×a×2

→ a ×4 = 44.72 ×44.72 = 1999.87 ≈ 2000

→ a = 2000/4

→ a = 500m/s²

As we know that Force is the product of mass and acceleration

→ F = m×a

Substitute the value we get

→ F = 50×500

→ F = 25000 N

∴ The force given by the net on the man is 25000 Newton.