# What is special about the 45 degree point for projectiles?

This is a question from a book for which I HAVE the solution to BUT I don't understand the concept. Briefly: "a mass is projected horizontally from a cliff with velocity V. 3 seconds later the direction of velocity is 45 degrees below horizontal. Taking g=10 and neglecting air resistance what is value of the initial projection velocity V in m/s." Now, I know the answer is 30. I understand that, neglecting air resistance, the horizontal component of velocity remains constant, but I what I don't understand is, when the direction of velocity is 45 degrees below the horizontal, why must the vertical component of velocity now numerically equal to the horizontal component? I know that sin 45 = cos 45 = 0.71 so wouldn't V at t=3 (and 45 degrees below) be Vcos45 and then wouldn't the vertical component at t=3 be (Vcos45)xsin45? I would greatly appreciate if you would just clarify the concept for me. Is this true for all horizontally-thrown, free falling objects then i.e. that at the 45 degree point the x and y components are equal?! How come?!

For a right angle triangle you can use SOH CAH TOA to find the relations between the angles and the lengths of the sides. If you want to find the relation between the horizontal and vertical components of the velocity you need to use the tangens and $\tan(45^\circ)=1$, thus the ratio between the horizontal and vertical components of the velocity is one.