During liftoff, a hot-air balloon accelerates upward at a rate of 3.2 m/s^2. The

Question

During liftoff, a hot-air balloon accelerates upward at a rate of 3.2 m/s^2. The balloonist drops an object over the side of the gondola when the speed is 18 m/s.

How long does it take to hit the ground? What I've been trying is using the x(t) = -1/2at^2 + V0t + X0 formula, where I would plug in g = 9.8 for a, and V0 = 0. Then I set x(t) to 0 and solved for t where I get: sqrt(2X0/g) = t. This is where I get stuck, I don't know what X0 is since the problem only gives limited information, how do I find X0? I tried plugging in 18 for it, but that's the velocity of where the object was dropped.
How long does it take to hit the ground? How long does it take to hit the ground?

Explanation / Answer

Initial velocity of the object is the same as that of the balloon and is in the direction of the balloon. u = 18 m/s Acceleration of the balloon a = 3.2 m/s^2 Distance covered by the balloon h = (18)^2/(2*3.2)                                                      = 50.625 m From kinematic relation                         h = -ut + (1/2)gt^2                   50.62 = -18t + 4.9t^2 4.9t^2 -18t - 50.62 = 0 Onsolving the above quadratic equation we get                time taken to reah the ground t = 5.53 s                                          

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