Question

In case of a crash the pilot of the plane jumped from it and opened hisparachute after falling through 40 m. Afterwards he decelerates at2 m/s2. If he reaches the ground with a velocity of 2 m/s, how longdid he stay in the air?

Answer 1

Given data

*The given distance is h = 40 m

*The given deceleration is a = 2 m/s^2

*The value of the acceleration due to gravirty is g = 9.8 m/s^2

*When the parachute reaches the ground, the final velocit

Case 1:

The formula for the velocity of the parachute when it covers a distance of 40 m under the free fall is given by the equation of motion as

`v^2=u^2+2gh`

Substitute the known values in the above expression as

`\begin{gathered} v^2=(0)^2+2(9.8)(40) \\ v=28\text{ m/s} \end{gathered}`

The time is calculated by the parachute as it covers a distance of 40 m under the free fall is given by the first equation of motion as

`\begin{gathered} v=u+at_1 \\ t_1=(v-u)/(a) \\ =(28-0)/(9.8) \\ =2.85\text{ s} \end{gathered}`

Case 2:

Parachutists decelerates at 2 m/s^2 and reach ground at a speed of 2 m/s

The formula for the distance traveled by the parachute is given by the equation of motion as

`\begin{gathered} v^2=u^2+2as \\ s=(v^2-u^2)/(2a) \end{gathered}`