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Question: A bullet is fired horizontally on a point target at \[100\;{\rm{cm}}\] with a speed of \[500\;{\rm{m...

A bullet is fired horizontally on a point target at 100  cm100\;{\rm{cm}} with a speed of 500  ms1500\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}. With what vertical distance, it will miss the target? [g=10  ms2]\left[ {g = 10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}}} \right]
A. 0.04  mm0.04\;{\rm{mm}}
B. 0.004  mm0.004\;{\rm{mm}}
C. 0.00004  m0.00004\;{\rm{m}}
D. 0.0045  m0.0045\;{\rm{m}}

Explanation

Solution

The above problem can be resolved using the fundamental equations of motion, especially the second equation of motion. First of all, the time taken by the bullet to hit the target is calculated, then using the value of time obtained from this equation, one can apply the second equation of motion. In the second equation of motion, the initial speed will be zero, as the body was initially at rest. Then by making the substitution of the numerical values, one can obtain the final result.

Complete step by step answer:
Given:
The horizontal distance of the point target is, d=100  cm×1  m100  cm=1  md = 100\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 1\;{\rm{m}}.
The speed of the bullet is, v=500  ms1v = 500\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}.
The value of gravitational acceleration is, g=10  m/s2g = 10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}.
The expression for the time take to hit the ground is,

v=dt t=dvv = \dfrac{d}{t}\\\ t = \dfrac{d}{v}

Solve by substituting the value in the above equation as,

t=dv t=1  m500  m/s t=2×103  st = \dfrac{d}{v}\\\ t = \dfrac{{1\;{\rm{m}}}}{{500\;{\rm{m/s}}}}\\\ t = 2 \times {10^{ - 3}}\;{\rm{s}}

Let the vertical distance be h, so that the bullet will miss the target.
Then the mathematical expression for the vertical distance is given by applying the second equation of motion as,
h=u×t+12gt2h = u \times t + \dfrac{1}{2}g{t^2}
Here, u is the initial speed of the bullet and its value is 0  m/s0\;{\rm{m/s}}.
Solve by substituting the values as,

h=u×t+12gt2 h=(0  m/s)×t+12×10×(2×103)2 h=12×10×2×103×2×103 h=0.00002  mh = u \times t + \dfrac{1}{2}g{t^2}\\\ \Rightarrow h = \left( {0\;{\rm{m/s}}} \right) \times t + \dfrac{1}{2} \times 10 \times {\left( {2 \times {{10}^{ - 3}}} \right)^2}\\\ \Rightarrow h = \dfrac{1}{2} \times 10 \times { {2 \times {{10}^{ - 3}}} }\times 2 \times {{10}^{ - 3}}\\\ \Rightarrow h = 0.00002\;{\rm{m}}

**Therefore, the vertical distance with which the bullet will miss the target is 0.00002 m and none of the options is correct.

Note: **
To resolve the given problem, one must remember the fundamental equations of motions and their mathematical formulas. These formulas are useful in resolving the typical problems in kinematics, that involves the motion in 1-D, motion in 2-D and the motion in 3-D.