A ball is projected from the ground with velocity v such that its ranege is maximum.

Answer

`ararrp;brarrq;crarrr;drarrs`

Explanation

The range of projectile is maximum.
`therefore theta=45^(@)`
and `H=(v^(2)sin^(2)45^(@))/(2g)=(v^(2))/(4g)`
(a) Using, `v_(y)^(2)-u_(y)^(2)=2a_(y)Y`
`v_(y)^(2)-(v^(2))/(2)=2(-g)(H)/(2)`
`=-g((v^(2))/(4g))=-(v^(2))/(4)`
`therefore v_(y)^(2)=(v^(2))/(4) rArr v_(y)=(v)/(2)`
`therefore (a)rarr(p)`
(b) At the maximum height, the ball has only horizontal velocity.
`therefore v_(x)=(v)/sqrt(2)`
`therefore (b)rarr(q)`
(c) Horizontal velocity remains the same but the vertical velocity gets reversed.
`therefore Deltav_(y)=(v)/sqrt(2)-(-(v)/sqrt(2))=(2v)/sqrt(2)=vsqrt(2)`
`therefore (c)rarr(r)`
(d) `vecv_(av)=(1)/(2)(vecv_(1)+vec_(f))`
`=(1)/(2)[(v)/sqrt(2)(hati+hatj)+(v)/sqrt(2)hati]`
`=(1)/(2)[(2v)/sqrt(2)hati+(v)/sqrt(2)hatj]`
`therefore |vecv_(av)|=(1)/(2)sqrt(((4v^(2))/(2)+(v^(2))/(2)))=(v)/(2)sqrt((5)/(2))`
`therefore (d) rarr(s)`