referring to an object being projected upwards at an angle ө i'm a phys student but i don't do math so i'm having trouble understanding how this equation has been simplified. Horizontal Range= vcosө x 2vsinө/g =v^2 2sinөcosө/g this is the line i don't understand : v^2 sin^2 ө/g where did the cosө go? is is a trig identity that cancels? also when considering upward motion is g taken as negative? if so, then why is the time taken to reach the maximum pt given by: t= v sinө/ g wouldn't it be v sinө/ -g isn't that the reason for the vertical distance travelled being given by; y= v sinө t- 1/2 gt^2 ??