Using equation of motion with constant acceleration, we can write:
v1=2gh1 and v2=2gh2
Now, required impulse
I=Δp=pf−pi
Given, M=0.1kg
Therefore,
I=Δp=0.1(2×9.8×5−(−2×9.8×10))
=0.1(72+14)≈2.39kgms−1
JEE Main 2024 — Physics Mechanics
A spherical body of mass 100g is dropped from a height of 10m from the ground. After hitting the ground, the body rebounds to a height of 5m. The impulse of force imparted by the ground to the body is given by: (given g=9.8ms−2)
Held on 30 Jan 2024 · Verified 6 Jul 2026.
4.32kgms−1
43.2kgms−1
23.9kgms−1
2.39kgms−1
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