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 air.
The Attempt at a Solution
Ok heres 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
This gives me
90.2=29.73 T+ .5T^2
So, Ill have T^3= (29.73+90.2)*2 ??
Giving me a time of 6.21 Seconds.
even though i use 29.73 m/s in the time equation 42.05 is my initial velocity correct??