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=−25 m,  uy=−5 sin 30°=−2.5 m/s (negative because vertical component acts downwards)(a) We have ay=−g=−10  m/s2     sy=uyt+1/2ayt2 −25=−2.5t−1/210t2So, therefore 10t2 + 5t - 50 = 0. On solving, we get is t = 2s, -2.5 s.The relevant time is 2s.(b) sx=uxt=5cos30°2=53 m.