From a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. The magnitude of average velocity during its ascent is

From a point on the ground a particle is projected with initial veloci

1.	From a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. The magnitude of average velocity during its ascent is
Answer» SOLUTION :Velocity of projection = u . Range is maximum `rArr THETA = 45^(@)` During time of ascent `rArr ` when H = `h_(mas)` `rArru_(x) = u _(x) = u * cos theta` Average velocity, `V_(A) = sqrt(V_(x)^(2) + V_(y)^(2))` `V_(x)`= Average velocity along x-axis ` = (u* cos theta + u cos theta)/(2)` `v_(x) = u * cos theta = u * 45^(@) = u //sqrt(2)to (1)` Average velocity along y-axis `= (u_(y) + v_(y))/(2)` `u_(y) = u sin theta = u // sqrt(2), v_(y) = 0 (at h_(max))` `:. v_(y) = (u//sqrt(2))/(2) = (u)/(2sqrt(2)) rarr (2)` Average velocity during time of ascent `sqrt((u/(sqrt(2)))^(2)+((u)/(2sqrt(2)))^(2))` Average velocity `= sqrt(u^(2)/(2)+(u^(2))/(4 xx 2))=sqrt((4u^(2) + u^(2))/(8))=(usqrt(5))/(2sqrt(2))`