Aman Sharma
Last Activity: 7 Years ago
We know that the distance covered vertically by the projectile is 15m so by equation of motion ( 2aysy = vy2 - uy2) we have s = 15m a=g u= usin60 thus (usin60)2 = 2g15m solving we get u2= 40g m let the value of g be 10ms--2 thus u2=400ms--1 by square rooting on both sides we get u=20ms--1