Question

The maximum height reached by a ball thrown at an amgle of 60'to the horizontal with an initial velocity 9.8 m/s is

Answer 1

By third equation of motion
( 9.8sin 60)^2 =2*9.8*h
h = 2*9.8/(9.8*9.8*3/4)=2*4/(3*9.8)=8/29.4 =80/294 m