Answer to Question #235927 in Classical Mechanics for Sridhar

Question

particle is projected vertically upwards from ground with speed 10m/s and second particle is at the same instant from a height of 6√3m directly above the partical with speed 10 m/s at an angle of projection 30

Answer 1

$$x_1=10\sqrt{\frac{2(6\sqrt{3})}{9.8}}=14.56\ m\\ x_2=\frac{(10)^2\sin{60}}{9.8}=8.84\ m$$