A man bails out form an aeroplane and after dropping vertically through a distance of 40 m, he opens the parachute and decelerates at 2 $m s^{- 2} .$ If he reaches the ground with a speed of 2 𝑚/𝑠, how much distance did he travel in air ?

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Solution

After falling 40 𝑚,
he attains speed $v^{2} = 0^{2} + 2 \times 9.8 \times 40,$
$v = 28 m / s$
When parachutist decelerates uniformly,
$u = 28 m / s, v = 2 m / s, a = - 2 m s^{- 2}$
Apply, $v^{2} = u^{2} + 2 a s,$
$2^{2} = 28^{2} - (2 \times 2 \times s)$
780 = 4s
We get, 𝑠 = 195 𝑚
Height at which parachutist bailed out = 195 + 40 = 235 𝑚

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A man bails out form an aeroplane and after dropping vertically through a distance of 40 m, he opens the parachute and decelerates at 2ms−2. If he reaches the ground with a speed of 2 𝑚/𝑠, how much distance did he travel in air ?

After falling 40 𝑚, he attains speed v2=02+2×9.8×40, v=28 m/s When parachutist decelerates uniformly, u=28 m/s,v=2 m/s,a=−2 ms−2 Apply, v2=u2+2as, 22=282−(2×2×s) 780 = 4s We get, 𝑠 = 195 𝑚 Height at which parachutist bailed out = 195 + 40 = 235 𝑚

A man bails out form an aeroplane and after dropping vertically through a distance of 40 m, he opens the parachute and decelerates at 2 $m s^{- 2} .$ If he reaches the ground with a speed of 2 𝑚/𝑠, how much distance did he travel in air ?