🚀 go-pugleaf

I have it now...


5.0 m = 4.0 m/s + 1/2 (10 m/s*s) t^2
t = 0.4 seconds to get to max height

max height is distance achieved at 0.4 seconds
y = V t + 1/2 g t^2
   = 4.0 m/s -.5(10 m/s*s)(0.4)^2
   = 1.6 m    -  0.8 m
   = 0.8 m

add that to initial height of 5.0 m gives 5.8 meters

ball reaches 5.8 meters, then starts falling

distance = 1/2 a t^2
5.8m = 1/2(10m/s*s)t^2
sq rt (5.8/5) = t
t = 1.077 s

total time is the two times added
0.4s + 1.08s = 1.48s


X component of velocity has to cover 8.0 meters in 1.48 seconds
8m/1.48s = 5.41 m/s



On Tue, 19 Feb 2008 22:43:10 -0800, John Dredla <Jdredla@hotmail.com>
wrote:

>The book has a companion website to practice with and try things out. 
>http://wps.aw.com/aw_young_geller_8/42/10907/2792221.cw/index.html
>
> I'm doing the target practice one (3.6)  and am completely stuck.
>
>Question 2
>
>I want to hit a target 8.0m away from the origin. I'm starting 5.0 m
>high.   My y component of velocity is given, 4.0 m/s.  I can play with
>the thing and guess an x component of velocity to be around 5.2 m/s,
>but I want to be able to calculate it.  I know I have to find the time
>it takes the ball to go up with 4.0 m/s velocity and then head back
>down to land on the ground.  Then I'll take that time and devide the
>8.0 m by that time to get x component.  They are using 10m/s^2  for
>gravity.
>
>How do you solve for t?
>