I have already figured that the magnitude of the objectsacceleration after it wa
ID: 1759957 • Letter: I
Question
I have already figured that the magnitude of the objectsacceleration after it was released is at 9.8m/s>2. Since I haveonly been given three gfigures to calculate the speed and none ofthem have distance to figure in any of the Free Fall equations, Iam left very confused as to what the possible answer might be.Could you please help. Thanks. During liftoff, a hot-air balloon accelerates upward at a rateof 2.6 m/s^2 . The balloonist drops an object over the side of thegondola when the speed is 18 m/s .How long does it take to hit the ground? I have already figured that the magnitude of the objectsacceleration after it was released is at 9.8m/s>2. Since I haveonly been given three gfigures to calculate the speed and none ofthem have distance to figure in any of the Free Fall equations, Iam left very confused as to what the possible answer might be.Could you please help. Thanks.Explanation / Answer
Use the acceleration of the balloon to figure out the heightof the balloon when the object is dropped. . time for balloon to reach theheight = final speed / acc = 18 / 2.6 = 6.923 sec . height of balloon when object isdropped = (1/2) a t2 = (1/2) *2.6 * 6.9232 = 62.306 meters . Now... this is the initial height of the dropped object. Andits initial speed is 18. And you can use the equation: . final position = initialposition + initial velocity * time + (1/2) at2 forthe object . 0 = 62.306 + 18 * t - (1/2) * 9.8 * t2 . 0 = 4.9t2 - 18 t - 62.306 quadratic equation. Use quadratic formula and youget... . t = 5.848 or -2.174 . Only the positive answer makes sense, so it takes 5.848 seconds for the object tohit the ground . 0 = 4.9t2 - 18 t - 62.306 quadratic equation. Use quadratic formula and youget... . t = 5.848 or -2.174 . Only the positive answer makes sense, so it takes 5.848 seconds for the object tohit the groundRelated Questions
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