Question

A hot-air balloon is drifting in level flight due east at 2.8 m/s due to a light wind. The pilot suddenly notices that the balloon must gain 28 m of altitude in order to clear the top of a hill 130 m to the east. How much time does the pilot have to make the altitude change without crashing into the hill? What minimum, constant, upward acceleration is needed in order to clear the hill? What is the horizontal component of the balloon’s velocity at the instant that it clears the top of the hill? What is the vertical component of the balloon’s velocity at the instant that it clears the top of the hill?

Answer 1

a. There is nothing suggesting that the balloon is accelerating horizontally, so we can assume that its horizontal speed is constant. The time before the balloon crash into the hill is simply the distance between the balloon and the hill divided by its velocity. Remember that velocity is simply the amount of distance that a object travels in a certain amount of time:

$t = \frac{130m}{2.8 m/s} = 46.43 s$

b. Know that you know the maximum amount of time that the balloon can take to gain 28m of altitude, the minimum acceleration can be found using the  equations constant acceleration motion:

$x = \frac{1}{2}at^2 + v_ot +x_0$

where a is the acceleration, v_o is the initial vertical velocity, 0 as the balloon is not moving vertically before starting to ascend. xo is the initial position, which we will give a value of 0m.

$x = \frac{1}{2}at^2\\ 28m = \frac{1}{2}(46.43s)^2a\\ a = \frac{2*28m}{(46.43s)^2} = 0.026 m/s^2$

c. As we said before, there isn't any kind of force that accelerates the balloon horizontally, therefore, we can consider that its horizontal velocity is constant and equal to 2.8m/s

d. Acceleration is the amount of change in velocity after a given amount of time. So, with the acceleration and the time we can fin the velocity:

$v_y = a_y*t = 0.026m/s^2*46.43s=1.206 m/s$