physics
A soccer ball is kicked with an initial speed of 10.8 m/s in a direction 24.9° above the horizontal. Calculate the magnitude of its velocity 0.310 s after being kicked. What was its direction? Calculate the magnitude of its velocity 0.620 s after being kicked. What was its angle relative to the horizontal (choose positive for above and negative for below)?

Vo = 10.8m/s[24.9o]
Xo = 10,8*cos24.9 = 9.8 m/s
Yo = 10.8*sin24.9 = 4.55 m/s
a. Y = Yo + g*t
Y = 4.55 + (9.8)*0.310 = 1.51 m/s.
V^2 = Xo^2 + Y^2
V^2 = 9.8^2 + 1.51^2 = 98.3
V = 9.9 m/s.
b. = tan A = Y/Xo = 1.51/9.8 = 0.15408
A = 8.76o = The direction.
c. Y = Yo + g*t
Y = 4.55 + (9.8)*0.620 = 1.53 m/s.
V^2 = Xo^2 + Y^2
V^2 = 9.8^2 + (1.53^2) = 98.38
V = 9.9 m/s.
tan B = Y/Xo = 1.53/9.8 = 0.15612
B = 8.87o = The direction.
NOTE: The negative velocity(1.53) means that the ball is falling and each
point during the rise time is repeated during the fall time.
posted by Henry
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