Recent content by mimsteel

Can anyone verify i worked that properly??

Ok here's where I got to. I used V^2=Vi^2+2a(90.2m) (296ft to 90.2m) This gives me a Vi of 42.05 m/s correct?? Now I take that 42.05m/s and get the X component and get 29.73 m/s and use formula X=ViT+.5T^2 This gives me 90.2=29.73 T+ .5T^2 So, Ill have T^3= (29.73+90.2)*2 ...

So, I need to solve for time before i can get the Vi correct?

Ya like i split it into X and Y components. And i have that. I just don't have a time. So I am not sure which formula to use from there. So like X=209.3 and the Y= 209.3 as well.

If someone could tell me what formula to use so i can start to try to figure this out. Not looking for a full solution just where to start off

Homework Statement A baseball was thrown at an angle of 45degrees above the horizonal, it traveled a horizontal distance of 296ft. and it was caught at the same level from which it was thrown. Neglect air resistance. A.) What was the ball's initial speed? B.) how long was the ball in the...

Homework Statement At one instant a bicyclist is 40.0m due east of a park's flagpole, going due south with a speed of 10.0m/s. Then 30.0s later, the cyuclist is 40.0m due north of the flagpole going due east with a speed of 10.0 m/s. For the cyclist in this 30.0s interval what are the A.)...