A small shell is fired from level…

A small shell is fired from level ground with speed 30 m s^-1 at an angle of 37^° above the horizontal (sin 37^°=0.6, cos 37^°=0.8). At the highest…

Numerical answer

A small shell is fired from level ground with speed

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

at an angle of

37^\circ

above the horizontal (

\sin37^\circ=0.6

\cos37^\circ=0.8

). At the highest point it explodes into two fragments of equal mass. One fragment has zero velocity immediately after the explosion and falls vertically; the other fragment continues horizontally. Take

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

and neglect air resistance. Find the distance, in metres, between the two landing points. Give the answer correct to two decimal places.

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Answer: 86.40

At projection

v_x=24\ \mathrm{m\,s^{-1}}

and

v_y=18\ \mathrm{m\,s^{-1}}

. The summit is reached after

1.8\ \mathrm{s}

at a horizontal distance

x_0=24\times1.8=43.2\ \mathrm{m}

and height

H=18^2/20=16.2\ \mathrm{m}

. There the velocity is purely horizontal,

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

. The explosion forces are internal, so horizontal momentum is conserved:

m(24)=\tfrac m2(0)+\tfrac m2v

, giving

v=48\ \mathrm{m\,s^{-1}}

for the second fragment. Both fragments then fall from

16.2\ \mathrm{m}

with no vertical velocity, taking

t=\sqrt{2H/g}=1.8\ \mathrm{s}

. The first lands at

x_0=43.2\ \mathrm{m}

and the second at

x_0+48\times1.8=129.6\ \mathrm{m}

, so the separation is

48\times1.8=86.40\ \mathrm{m}

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Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake