A) 30 m
B) 40 m
C) 20 m
D) 10 m
Correct Answer: B
Solution :
Bullet will collide with piece of wood at \[t=1\]second at \[t=1\] second Velocity of piece of wood \[=u=0+10\times 1\] \[=\text{ }10\text{ }m\text{ }m/s\] Velocity of bullet = v \[=\text{ }100-10\times 1=90\text{ }m/s\uparrow \] from momentum conservation \[\left( {{p}_{i}}={{p}_{f}} \right)\] \[{{p}_{i}}=90\times 0.02-0.03\times 10=1.8-0.3=1.5\] \[{{p}_{f}}=\left( 0.03+0.02 \right)\text{ }{{v}_{f}}=0.05\text{ }{{v}_{f}}\] \[\therefore \,\,{{v}_{f}}\,=\,\frac{1.5}{0.05}\,=\,30\,m/s\] Height from the point of collision H \[=\,\,\frac{30\times 30}{2\times 10}\,\,\,=\,\,\,45\,m\] At time of collision system is h distance below top of tower where \[h=\frac{1}{2}\times 10\times {{1}^{2}}=5\] \[\therefore \] Height above tower \[=\text{ }45-5=40\] metreYou need to login to perform this action.
You will be redirected in
3 sec