A projectile shot at angles of `45^(@)` above the horizontal strikes a buliding 30 m away at a point 15 m above the point of projection. Find : (a) the speed of projection . (b) the magnitude and direction of velocity of projectile when it strikes the building .

Explanation

Let "u" is the speed of projection .
(a) Let P be the point on the building where projectile hits it .
Taking point of projection as origin, coordinates of P are (30,15)
Using the equation of trajectory .
`y = x tan theta -(gx^(2))/(2u^(2) cos^(2) theta) rArr 15 = 30tan 45^(@) - (g(30)^(2))/(2u^(2) cos^(2) 45^(@)) rArr u = 24.2ms^(-1)`
(b) At `p, v_(x) = u_(x) = 24.2 cos 45^(@) = 17.11 ms^(-1)`
`v_(y)^(2)= u_(y)^(2) + 2a_(y)s_(y) rArr v_(y)^(2) = u^(2) sin^(2) 45^(@) - 2g(15)`
`v_(y)^(2) = (60g) (0.5) - 30 g =0 rArr v_(y) =0`
At P , projecticle is at its highest point and hence moving horizontally.