A ball launched from ground level lands 2.15 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

Answer 1

u = constant horizontal speed = 50/2.5 = 20 m/s

v = vi - 9.8 t
at top of arc, t = 2.5/2 = 1.25 seconds and v = 0
so at top
0 = vi - 9.8 (1.25)
so
vi = 12.25

so speed at start = sqrt (20^2 + 12.25^2)
= sqrt(400+ 150) = sqrt 550 = 23.5 m/s

tan angle = 12.25/20
so
angle = 31.5 degrees