A body is projected with a velocity `vecv =(3hati +4hatj) ms^(-1)` The maximum height attained by the body is: `(g=10 ms^(-2))`

Answer

A, B, C, D

Explanation

[A-p, B-q, C-r, D-s]
A-p Range is maximum, when the angle of projection is `45^(@)`
`H=v^(2)/(2g) sin^(2) 45=v^(2)/(4g) .........(1)`
Velocity, at half of the maximum height is v'.
`v_(y)^(2)=v^(2) sin^(2) 45-2g H/s=v^(2)/2-v^(2)/4 rArr v_(y)=v/2 rArr v' =sqrt(3/2)v`
B-q Velocity at the maximum height is v'
`v_(x)=v cos 45 rArr v_(x)=v/sqrt2 rArr v'= sqrt(v_(x)^(2)+v_(y)^(2))=v/sqrt2`
[Because vertical component of velocity is zero at the highest point]
C-r Projection velocity
At projection point `v_(1)=v cos 45hati+v sin 45 hatj` At the point, when the body strikes the ground `vecv f=v cos 45 hati-v sin 45 hatj , trianglev=vecv_(f)+(-vecv_(i))=2v sin 45 (-j), |trianglevecv|=2v sin 45=vsqrt2`
D-s Average velocity `=("Total displacement")/("Total time")`
Diplacement `=sqrt((R/2)^(2)+H^(2)), v_(av) =sqrt((R^(2))/(4)+H^(2))/((v sin theta)/(g))=sqrt(R^(2)+4H^(2))/((2v sin theta)/(g))`
`v_(av)=(sqrt(((v^(2))/(g)))+4((v^(2))/(4g))^(2))/((sqrt2v)/(g))=((v^(2))/(g)sqrt(+1/4))/(vsqrt2)/(g)=(v^(2)sqrt5)/(2vsqrt2)=v/2 sqrt(5/2)`