Two paper screens A and B are separated by a distance of 100m. A bullet pierces A and then B. the hole in B is 10 cm below the hole in A. if the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is

Explanation

Assume that the bullet hits the screen A with velocity u and pierces the screen B after time t.
`therefore` Horizontal distance, `PQ=ut=100m` or `t=(100)/(u)`
Vertical distance, `QR=(1)/(2)"gt"^(2)=0.1` or `(1)/(2)g((100)/(u))^(2)=0.1`
`(1)/(2)gxx(10000)/(u^(2))=0.1 rArr u^(2)=(9.8xx(100)^(2))/(2xx0.1)=49xx(100)^(2)rArr u=700 ms^(-1)`
Alternative method: `R=usqrt((2H)/(g)) R=100m, H=0.1 m,g=9.8 m//s^(2)`
`100=usqrt(2xx0.1)/(9.8)=usqrt((1)/(49))=(u)/(7)rArru=7xx100=700 m//s`