A stone is projected with a velocity `20sqrt2 m//s` at an angle of `45^(@)` to the horizontal.The average velocity of stone during its motion from starting point to its maximum height is

Options

A. `5sqrt5 ms ^(-1)`
B. `10sqrt5 ms ^(-1)`
C. `20 ms^(-1)`
D. `20sqrt5 ms ^(-1)`

Answer

B

Explanation

When projectile is at A, then
`OC = R/2=1/2(u^2)/(g)sin2theta`
`=1/2 xx ((20sqrt2)^2)/10sin2 xx 45^@ = 40 m `
`AC = H = (u^2sin^2 theta)/(2g)=((20sqrt2)^2)/(2xx10)sin^2 45^@=20m`
`:.` Displacement , `OA = sqrt(OC^2 +CA^2)=sqrt(40^@+20^@)`
Time of projectile from O to A
`=1/2((2usintheta)/g)=(usin theta)/g=((20sqrt2)sin 45^@)/10=2s `
`:.` Average velocity = `("displacement")/(" time") =(sqrt(40^2+20^(2)))/2=10sqrt5ms^(-1)`