Asked by Tsunayoshi
A rock is thrown from the ground at a speed of 22m/s and an angle of 55 degrees above the horizontal. Find the velocity and height of the rock after 2 seconds.
Answers
Answered by
Henry
Vo = 22m/s[55o]
Xo = 22*cos55 = 12.62 m/s.
Yo = 22*sin55 = 18.02 m/s.
h = Yo*t + 0.5g*t^2
h = 18.02*2 - 4.9*2^2
Y = = Yo + g*t
18.02 - 9.8*2 = -1.58 m/s. = Vertical
component of velocity. The negative sign means the rock is falling.
Xo = 22*cos55 = 12.62 m/s.
Yo = 22*sin55 = 18.02 m/s.
h = Yo*t + 0.5g*t^2
h = 18.02*2 - 4.9*2^2
Y = = Yo + g*t
18.02 - 9.8*2 = -1.58 m/s. = Vertical
component of velocity. The negative sign means the rock is falling.
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