Asked by Viviana
A stone is thrown up at an angle of 30degrees to the horizontal with a speed of 20metres per second from the edge of a cliff 15m above sea level, so that the stone lands in the sea. Find how long the stone is in the air and how far from the base of the cliff it lands.
Answers
Answered by
Henry
Vo = 20m/s[30o].
Xo = 20*Cos30 = 17.3 m/s.
Yo = 20*sin30 = 10 m/s.
Yf = Yo + g*Tr.
Yf = 0.
g = -9.8 m/s^2.
Tr = Rise time = ?.
h = 15 + Yo*Tr + 0.5g*Tr^2 = Meters above sea level.
g = -9.8 m/s^2.
h = 0.5g*Tf^2.
g = 9.8 m/s^2.
Tf = Fall time = ?.
Tr + Tf = Time in air.
Dx = Xo*(Tr+Tf) = Hor. distance from base.
Xo = 20*Cos30 = 17.3 m/s.
Yo = 20*sin30 = 10 m/s.
Yf = Yo + g*Tr.
Yf = 0.
g = -9.8 m/s^2.
Tr = Rise time = ?.
h = 15 + Yo*Tr + 0.5g*Tr^2 = Meters above sea level.
g = -9.8 m/s^2.
h = 0.5g*Tf^2.
g = 9.8 m/s^2.
Tf = Fall time = ?.
Tr + Tf = Time in air.
Dx = Xo*(Tr+Tf) = Hor. distance from base.
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