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

The pilot has approximately 46.43 seconds to gain 28 m in altitude to clear the hill. The minimum constant upward acceleration needed is approximately 0.026 m/s². The horizontal component of the balloon's velocity at clearance will be 2.8 m/s, while the vertical component will be approximately 1.21 m/s.

Step-by-step explanation:

The time the pilot has to make the altitude change without crashing into the hill can be found using the horizontal velocity and the distance to the hill. Since the balloon drifts horizontally at 2.8 m/s and needs to cover 130 m, the time (t) it will take can be calculated as:

t = distance / horizontal velocity = 130 m / 2.8 m/s = 46.43 seconds.

To find the minimum constant upward acceleration (a) needed to clear the hill, we use the kinematic equation:

s = ut + (1/2)at2

Where s is the vertical displacement (28 m), u is the initial vertical velocity (0 m/s), and t is the time calculated above. Rearranging for a gives:

a = 2s / t2 ≈ 2(28 m) / (46.43 s)2 ≈ 0.026 m/s2.

As the horizontal velocity is not affected by the vertical motion in the absence of air resistance, the horizontal component of the balloon's velocity when it clears the top of the hill will remain 2.8 m/s.

To find the vertical component of the velocity at the instant it clears the top of the hill, we can use the equation:

vf = u + at

Where vf is the final vertical velocity, u is the initial vertical velocity, a is the acceleration, and t is the time taken. Substituting the known values gives:

vf = 0 m/s + (0.026 m/s2)(46.43 s) ≈ 1.21 m/s.

Answer 2

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 = (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 = (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 = (1)/(2)at^2\\ 28m = (1)/(2)(46.43s)^2a\\ a = (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`