x=40=v0t
y=20= 12gt2
Get: t=2s, v0=20m/s, the direction is horizontal to the right,
A=g= 10m/s2, and the direction is vertical downward.
(2)vy=gt=20m/s
Vt = v2x+v2y = 202+202 = 202 = 202m/s.
Tan θ = Vyvx = 2020 = 1。 That is, the direction is inclined downward at an angle of 45 degrees with the horizontal plane.
(3)mgcos450=mv2tr
r = 2v2t2g = 2×(202)22× 10 = 802m。
Answer: (1) Study the curvature radius of this curve with the model of quasi-flat throwing motion or flat throwing motion;
The speed of the starting point (coordinate origin) is 20m/s, and the direction is horizontal to the right.
The acceleration g= 10m/s2, and the direction is vertical downward.
(2) The speed of the point with coordinate (40, -20) is 202m/s, and the direction is inclined downward at an angle of 45 degrees with the horizontal plane.
(3) The radius of curvature r of the point with the coordinate (40, -20) is 802m. ..