Posted by **lora** on Monday, June 18, 2012 at 1:18pm.

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?

- physical science -
**Elena**, Monday, June 18, 2012 at 2:40pm
Upward motion:

v=vₒ-g•t

at the top point

0= vₒ-g•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.

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