A body is projected vertically upwatds at time t=0 and is it seen at a height H at time `t_1` and `t_2` second during its flight. The maximum height attainet is (g is acceleration due to garavity).

Options

A. `g/4(t_1+t_2)^2`
B. `g((t_1+t_2)/4)^2`
C. `2g((t_1+t_2)/4)^2`
D. `g/4(t_1t_2)`

Explanation

`y=10x-(5/9)x^2`
on comparing with the standard equation,
`y=xtantheta-g/(2u^2cos^2theta).x^2,`
`tantheta=10,g/(2u^2cos^2theta)=5/9)`
The expression for range of projectile,R is given as,
`R=(2u^2sintheta.costheta)/g`
`=(2u^2tantheta,cos^2theta)/(g ) ( :.tantheta=sintheta/costheta)`
`9/5tantheta ( :'g/(2u^2cos^2theta)=5/9)
9/5xx10`
`:.r=18m`.