A projectile is given an initial velocity of 5 m/s at an angle 30° below horizontal from the top of a building 25 m high. Find (a) the time after which it hits the ground; (b) the distance from the building where it strikes the ground. (g = 10 m/s2)

The projectile is thrown from point O and lands at point A on the ground.From O to A : sy = -25m, uy = -5 sin30o = -2.5m / s (negative because vertical component acts downwards)(a) We have ay=−g=−10m/s2 sy=uyt+(1/2)ayt2−25=−2.5t−(1/2)(10)t2So, therefore 10t2 + 5t - 50 = 0. On solving, we get is t = 2s, -2.5 s.The relevant time is 2s.(b) sx=uxt=5cos⁡30∘2=53m.