A bullet on penetrating 30 cm into its target loses its velocity by 50- What additional distance will it penetrate into the target before it comes to rest-30 cm10 cm20 cm5 cm

Question

Toppr Admin · Accepted Answer

Given- S-30-xA0-cm-30-xD7-10-x2212-2m and loss in velocity is 50-Equation of motion-v2-u2-2asHere- u24-u2-2a-xD7-30-xD7-10-x2212-2 - -i-0-u24-2a-xD7-x - -ii-On solving Eqs- -i- and -ii- we getx-10-xA0-cm

A bullet on penetrating $30 c m$ into its target loses its velocity by $50$ %. What additional distance will it penetrate into the target before it comes to rest?
$30 c m$
$10 c m$
$20 c m$
$5 c m$

Given: $S = 30 c m = 30 \times 10^{- 2} m$ and loss in velocity is $50$ %
Equation of motion:
$v^{2} = u^{2} + 2 a s$
Here, $\frac{u^{2}}{4} = u^{2} + 2 a \times 30 \times 10^{- 2}$ ...... (i)
$0 = \frac{u^{2}}{4} + 2 a \times x$ ......... (ii)
On solving Eqs. (i) and (ii), we get
$x = 10 c m$

A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distance will it penetrate into the target before it comes to rest?30 cm10 cm20 cm5 cm

Given: S=30 cm=30×10−2m and loss in velocity is 50%Equation of motion:v2=u2+2asHere, u24=u2+2a×30×10−2 ...... (i)0=u24+2a×x ......... (ii)On solving Eqs. (i) and (ii), we getx=10 cm

A bullet on penetrating $30 c m$ into its target loses its velocity by $50$ %. What additional distance will it penetrate into the target before it comes to rest?
$30 c m$
$10 c m$
$20 c m$
$5 c m$

Given: $S = 30 c m = 30 \times 10^{- 2} m$ and loss in velocity is $50$ %
Equation of motion:
$v^{2} = u^{2} + 2 a s$
Here, $\frac{u^{2}}{4} = u^{2} + 2 a \times 30 \times 10^{- 2}$ ...... (i)
$0 = \frac{u^{2}}{4} + 2 a \times x$ ......... (ii)
On solving Eqs. (i) and (ii), we get
$x = 10 c m$