A parachutist jumps freely from an aero plane for 10 sec, and then his parachute opens out. Now we descends with net retardation of 2.5m/s². If he bails out of plane at height of 2495m and g = 10m/s², his velocity on reaching the ground will be 5x m/s, then x =
Physics · Mixed Chapter
JEE Mains

Question

A parachutist jumps freely from an aero plane for 10 sec, and then his parachute opens out. Now we descends with net retardation of 2.5m/s². If he bails out of plane at height of 2495m and g = 10m/s², his velocity on reaching the ground will be 5x m/s, then x =

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: During first 10 s of free fall, distance s₁=½gt²=500 m and downward speed v₁=gt=100 m/s. Step 2: Remaining vertical distance =2495−500=1995 m. Opening the parachute makes the net acceleration upward, a=−2.5 m s⁻². Step 3: v²=v₁²+2as=10000−5×1995=25. Hence impact speed v=5 m/s. Step 4: Since 5x=5, x=1. Answer: 1.