a ball thats 1 kg is thrown at 30 m/s into the air. how high did the ball travel then?

Answer 1

Mewthode1.
Kinetic energy on the botom = potential energy at the top
Ke = Pe
Ke=0.5 m V^2
Pe= mgh

then
h=0.5 m V^2/mg
h= 0.5 V^2 / g
h= 0.5 (30)^2/ 9.81
h=45.9m

Method #2

h=0.5gt^2
V=gt
t=v/g and again we have
h=0.5 V^2/g=45.9m