A projectile is launched from the ground at a speed of 40ms^-1 at an angle of 30 degrees to the horizontal, where does it land? What is the highest point the projectile reaches?

Since the SUVAT equations of motion act independently in the horizontal and vertical directions we can use the vertical equations to find the total time (the time when total vertical displacement is zero) and then calculate how far the partical has moved horizontally in that time.

Vertical:

s=ut+1/2at^2 ------> 0 = 40sin(30) t -0.59.8t^2----> t=4.08s

Horizontal:

s=ut+0.5at^2 -------> x=40cos(30)*4.08  [no horizontal acceleration] =141.3m

To find the apex of the flight, we use a second SUVAT equation, noting that the instantaneous velocity at the maximum is entirely horizontal.

Vertically:

v^2=u^2+2aS------>0=(40sin(30))^2-29.8H------------->H=400/19.6=20.4m