Question

A parachutist, after bailing out, falls 50 m without air resistance. When parachute opens, it
retards at 2 2ms . He reaches the ground with a speed of 1 3ms . At what height, did he bail
out

Answer 1

Answer:

• Initial Height = 50 m
• Air resistance = 0
• Retardation = 2 m/s²
• Final Velocity = 3 m/s
• Height at which he bailed out = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

• We shall first find the initial velocity which will be given by √2gh

$\displaystyle\sf :\implies u = \sqrt{2gh}$

$\displaystyle\sf :\implies u = \sqrt{2\times 10\times 50}$

$\displaystyle\sf :\implies u = \sqrt{1000}$

$\displaystyle\sf :\implies u = 10\sqrt{10}$

• Now we shall use the third Equation of motion as we have the initial, the final velocities along with the acceleration of the body

$\displaystyle\sf :\implies v^2-u^2 = 2as$

$\displaystyle\sf :\implies 3^2-(10\sqrt{10})^2 = 2\times (-2) \times s$

$\displaystyle\sf :\implies 9 - (100\times 10) = -4s$

$\displaystyle\sf :\implies 9-1000 = -4s$

$\displaystyle\sf :\implies -991 = -4s$

$\displaystyle\sf :\implies \dfrac{-991}{-4} = s$

$\displaystyle\sf :\implies\underline{\boxed{\sf Distance = 247.75 \ m}}$

$\displaystyle\therefore\:\underline{\textsf{Height at which he bail out is \textbf{ 247.75 m }}}$