JEE Main 2016 · Physics · Kinematics and Laws of Motion · easy

A ball A is dropped from rest from the top of an 80 m tower. One second later a ball B is thrown vertically upward from the ground, directly below…

Single correct option

A ball

A

is dropped from rest from the top of an

80\ \mathrm{m}

tower. One second later a ball

B

is thrown vertically upward from the ground, directly below, with a speed of

40\ \mathrm{m\,s^{-1}}

. Take

g=10\ \mathrm{m\,s^{-2}}

. The height above the ground at which the two balls meet is

A. $60\ \mathrm{m}$
B. $43.75\ \mathrm{m}$
C. $48.75\ \mathrm{m}$
D. $55\ \mathrm{m}$

Show answer and solution

Answer: (C) $48.75\ \mathrm{m}$

Measure time

t

from the release of

A

. Then

y_A=80-5t^2

and, since

B

starts a second later,

y_B=40(t-1)-5(t-1)^2

. Setting

y_A=y_B

80-5t^2=40t-40-5t^2+10t-5

, so

80=50t-45

and

t=2.5\ \mathrm{s}

. Substituting,

y=80-5(2.5)^2=80-31.25=48.75\ \mathrm{m}

. As a check,

y_B=40(1.5)-5(1.5)^2=60-11.25=48.75\ \mathrm{m}

. The quadratic terms cancel because both balls have the same acceleration, which is why the equation turns linear. The common mistake is to forget the one-second delay, which gives

60\ \mathrm{m}

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