I actually figured it out today right after class (guess it got me in the right type of mode to do it). Sometimes I just don't look at them the way I should be. But what I ended up doing is plugging in 19 m/s for v, -9.8 for g (a) and 0.97 in for t into the equation
v = v0 + at
This gave me...
I used to have one but I have no clue where it is at anymore. We aren't aloud to have them in this class so I have kinda lost track of it while I was moving.
I know I should be able to do this problem but for some reason it's not clicking with me...please help!
Homework Statement
a ball is thrown vertically upward with an initial speed of 19 m/s. Then, .97 s later, a stone is thrown straight up (from the same inital height as the ball) with an...
Mjsd...that makes sense to me except I don't understand why 's' is equal to '1'?? Also, in the s = 0.5 (u+v)t equation I would think that acceleration should be used somewhere in the equation since that is how fast it is decelerating.
My original idea was the use the equation I mentioned above...
A motorist is traveling at 18 m/s^2 when he sees a deer 38 m ahead. If the maximum negative acceleration is -4.5 m/s^2, what is the reaction time (delta t) of the motorist that will allow him to avoid hitting the deer? Answer in units of 's'.
I tried using x(t)=x0+v0t+1/2at^2 and it didn't...