Question

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming to rest. what was the deceleration experienced by the survivor? Use g = 9.8 m/s2 Calculate to one decimal.

Answer 1

The deceleration is
`a = - 76.27 m/s^2`

Step-by-step explanation:

From the question we are told that

The height above firefighter safety net is
`H = 14 \ m`

The length by which the net is stretched is
`s = 1.8 \ m`

From the law of energy conservation

`KE_T + PE_T = KE_B + PE_B`

Where
`KE_T` is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and
`PE_T` is the potential energy of the before jumping which is mathematically represented at

`PE_T = mg H`

and
`KE_B` is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

`KE_B = (1)/(2) m v^2`

and
`PE_B` is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

`mgH = (1)/(2) m v^2`

=>
`v = √(2 gH )`

substituting values

`v = 16.57 m/s`

Applying the equation o motion

`v_f = v + 2 a s`

Now the final velocity is zero because the person comes to rest

So

`0 = 16.57 + 2 * a * 1.8`

`a = - (16.57^2 )/(2 * 1.8)`

`a = - 76.27 m/s^2`