Respuesta :
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
[tex]v^2=u^2+2gh[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v^2=(0)^2+2(9.8)(40) \\ v=28\text{ m/s} \end{gathered}[/tex]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
[tex]\begin{gathered} v=u+at_1 \\ t_1=\frac{v-u}{a} \\ =\frac{28-0}{9.8} \\ =2.85\text{ s} \end{gathered}[/tex]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
[tex]\begin{gathered} v^2=u^2+2as \\ s=\frac{v^2-u^2}{2a} \end{gathered}[/tex]