Question

A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 21.0 m/s from a height of 9.0 m.
(a) How high does the ball rise from its original position?

(b) How long does it take to reach its highest point?

(c) How long does the ball take to hit the ground after it reaches its highest point?

(d) What is the ball's velocity when it returns to the level from which it started?

Answer 1

a) hmax=vy^2/2g
so 21^2/(2*9.8)= 22.5 m

b) t=sqroot 2h/g
sqrt of 2*22.5 /9.8 = 4.59s

c) you can use kinematics here
h=9+22.5 = 31.5
vi=0
a=9.8
you can use
d=1/2 gt^2 to find time
31.5=(.5*9.8)t^2
t=2.535s

d) then use vf =vi+at
where vf=? vi=0 a=g=9.8 t=2.535

hope that helped