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More physics help?
A person stands on the edge of a 3.3 m tall cliff. The person then throws a ball at a speed of 12.5 m/s at an angle of 47 degrees. What is the:
A: Max height the ball reaches?
B: The total time in the air?
C: Max distance the ball reaches?
If you have any advice, I would greatly appreciate it!
1 Answer
- oldprofLv 72 years agoFavorite Answer
Max height H = h + Uy^/2g = h + (U sin(theta))^2/2g; where h = 3.3 m, U = 12.5 m/s, and theta = 47 deg. g = 9.81 m/s^2. ANS a you can do the math.
T = Tu + Td = Uy/g + sqrt(2H/g) = (U sin(theta))/g + sqrt(2H/g) = total flight time. ANS b. Where Tu is the time to reach max height and Td is time to fall from that height.
X = Ux T = U cos(theta) T = max distance assuming no air drag ANS c.
That's my advice, do the math.