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Q. A body dropped from a height h with initial velocity zero, strikes the ground with a velocity $3 \,m/s$. Another body of same mass dropped from the same height h with an initial velocity of $4 \,m/s$. The final velocity of second mass, with which it strikes the ground is

AIPMTAIPMT 1996Motion in a Straight Line

Solution:

Initial velocity of first body $(u_1) = 0 ;$
Final velocity $(v_1) = 3\, m/s$ and initial velocity of
second body $(u_2) = 4\, m/s$
height $(h)=\frac{v_1^2}{2g}=\frac{(3)^2}{2 \times9.8}=0.46\, m.$
Therefore velocity of the second body,
$V_2=\sqrt{u_1^2+2gh}=\sqrt{(4)^2+2 \times9.8 \times 0.46}{}=5\, m/s.$