An object is projected so that it just clears two walls of height 7.5 m and with separation 50m from each other. If the time of passing between the walls is 2.5s, the range of the projectile will be (g=`10m//s^(2)`)

Options

A. 35m
B. 70m
C. 140m
D. 57.5m

Answer

b

Explanation

`(ucos theta)xx2.5=50`
`u_(x)=u cos theta=20` ..........(i)
`(2v_(y))/(g)=2.5 rArrv_(y)=12.5` ............(ii)
Now, `v_(y)=u_(y)-gxxt`
and `(v_(y))^(2)=(u_(y))^(2)-2gh`
`(u_(y))^(2)=(v_(y))^(2)+2gh`
`(u_(y))^(2)=(12.5)^(@)+2xx10xx7.5` .............(iii)
Hence `u=sqrt((u_(x))^(2)+(u_(y))^(2))`