physical science
posted by lora on .
if a ball is thrown straight up at a speed of 34.7m/s how long in seconds will it continue to rise before it starts to fall back down?

Upward motion:
v=vₒg•t
at the top point
0= vₒg•t1,
t1= vₒ/g =34.7/9.8=3.54 s.
h= vₒ•t1 –g•t1²/2=34.7•3.54  9.8•3.54²/2 =61.5 m
Downward motion
h=g•t2²/2,
t2=sqrt(2•h/g) =
=sqrt(2•61.5 /9.8) = 3.54 s.
Time (upward) =time (downward)!
t =2•t1=2•3.54 =7.08 s.