Asked by joseph
A group of engineering students constructs a slingshot to launch water balloons. The angle of launch is 30° and the initial velocity of the balloon is 14 m/s. The students launch a balloon from the roof of a building and the balloon lands 42 m away from the building. How tall is the building?
Answers
Answered by
Henry
Tr=(Y-Yo)/g = (0-14sin30)/9.8 = 0.71 s.
Range = Xo*(Tr+Tf) = 42 m.
14*cos30*(0.71+Tf) = 42
12.12*(0.71+Tf) = 42
8.61 + 12.12Tf = 42
12.12Tf = 42 - 8.61 = 33.39
Tf = 2.75 s.
h = Vo*t + 0.5g*t^2.
h = 0 + 4.9*2.75^2 = 37.1 m.
Range = Xo*(Tr+Tf) = 42 m.
14*cos30*(0.71+Tf) = 42
12.12*(0.71+Tf) = 42
8.61 + 12.12Tf = 42
12.12Tf = 42 - 8.61 = 33.39
Tf = 2.75 s.
h = Vo*t + 0.5g*t^2.
h = 0 + 4.9*2.75^2 = 37.1 m.
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