A human cannonball is launched from a cannon at 26.4 m/s at 20.4 degrees above the horizontal. (Assume he/she lands in a net the same height as the cannon.) How high in the air does he/she go?

Answer 1

viy=26.4*sin20.4

at the top, vyfinal=0
vyfinal=viy-g*t
solve for time t, then h=viy-1/2 g t^2