Asked by Camden
A ball bounces off the edge of a table at an initial velocity of 1.5 m/s at an angle of 30 degrees above the table, falling to the floor 75 cm below. How long will the ball be in the air?
Answers
Answered by
Henry
Vo = 1.5m/s[30o].
Xo = 1.5*Cos30 = 1.30 m/s.
Yo = 1.5*sin30 = 0.75 m/s.
Y = Yo + g*Tr.
0 = 0.75 - 9.8Tr, Tr = 0.077 s. = Rise time.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 0.75 + 0.75*0.077 - 4.9*0.077^2 = 0.779 m. Above the floor.
h = 0.5g*Tf^2.
0.779 = 4.9Tf^2, Tf = ?. = Fall time.
Tr+Tf = Time in air.
Xo = 1.5*Cos30 = 1.30 m/s.
Yo = 1.5*sin30 = 0.75 m/s.
Y = Yo + g*Tr.
0 = 0.75 - 9.8Tr, Tr = 0.077 s. = Rise time.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 0.75 + 0.75*0.077 - 4.9*0.077^2 = 0.779 m. Above the floor.
h = 0.5g*Tf^2.
0.779 = 4.9Tf^2, Tf = ?. = Fall time.
Tr+Tf = Time in air.
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