A hot air balloon decends toward the ground with a velocity of (1.6 m/s). A cham

Question

A hot air balloon decends toward the ground with a velocity of (1.6 m/s). A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (4.0 m/s) relative to the balloon. When opened, the bottle is 6.2 m above the ground. (Neglect air resistance.) (a) What is the initial velocity of the cork, as seen by an observer on the ground? = ( m/s) + ( m/s) (b) What is the speed of the cork, and its initial direction of motion, as seen by the same observer? Speed m/s Direction

Explanation / Answer

Assuming you answered the previous questions correctly. C) the max heigh will be from where its opened at. 6.2 m above the ground. This is because its being shot horizontally, so there is no angle of inclination and we are neglecting air resistance. D) The time it takes for it to hit the ground we can use the equation: ?y = vot + (1/2)at^2 We dont want to use x because then wed have to find final velocity, or if you tried using the kinematic ?x = then youd just get stuck solving for ?x (-6.2) = (-2.4)(t) + (1/2)(-9.81)(t^2) Note this will be a quadratic: 0 = (-4.905t^2) - (2.4t) + 6.2 Just plug and chug using quadratic formula. Neglect the negative because we do not really care about negative time. 0.905947

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