Recent content by SayedD

In a certain reference frame it is right?

My method is silly but here is my attempt Initially we know that the kinetic energy is k = ##1/2mv^2## and the final kinetic energy at the moment before splitting is equal to ##(k + 3k/16)mv^2_1 = (19k/16)mv^2_1##. Substituting the value k to the second equation gives us ##v_1 =...

So using the projectile equation I got ##a\sin (\alpha + \beta) = v\sin (\alpha + \beta)t + \frac{g\sin \beta t^2}{2}## To find t we use the other projectile equation: Since the horizontal velocity must be 0 for it to hit the ground perpendicularly we set the equation to be: ##0 = v\sin (\alpha...