Question

A ball with a mass of 250g is thrown straight upwards with a speed of 20m/s. It is caught on its way
down at a point 5m above the original point is was thrown.
I. How fast was it going when it was caught?
II. How long did the trip take?
III. What was the maximum height attained?
IV. What was the potential energy at the maximum height attained

Answer 1

I.
v=√(〖20〗^2-2(10)(5))=17 m/s.
II.
t=2V/g-(V-v)/g=(V+v)/g=(17+20)/10=3.7 s
III.
h=〖20〗^2/(2(10))=20 m.
IV.
E=mgh=(0.25)(10)(20)=50 J.