Question

Movies and TV shows sometimes portray a person being thrown backwards a sizable distance as a result of being struck by a bullet. Calculate the maximum distance that a 60.3-kg object would fly backwards if it is struck by a 10.0-gram projectile moving with a velocity 389 m/s which then completely embeds itself in the object. Assume that the angle of this perfectly inelastic collision is such that the combined mass is launched at an angle of 45 degrees above the horizontal, and neglect the effects of air resistance. Based on this calculation, evaluate the realism (or lack thereof) of the TV/movie portrayal.

Answer 1

Answer:

 R = 4.24 x 10⁻⁴ m

Explanation:

given,

mass of the person = 60.3-kg

mass of the bullet = 10 gram = 0.01 Kg

velocity of bullet = 389 m/s

angle made with the horizontal = 45°

using conservation of momentum.

M v  + m u  = ( M + m ) V

60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V

$V = \dfrac{3.89}{60.31}$

$V = \dfrac{3.89}{60.31}$

V = 0.0645 m/s

for calculation of range

$R = \dfrac{V^2sin 2 \theta}{g}$

$R = \dfrac{0.0645^2sin 2 (45^0)}{9.8}$

     R = 4.24 x 10⁻⁴ m

the distance actor fall is  R = 4.24 x 10⁻⁴ m