A hot-air balloon is drifting in level flight due east at 2.9 m/s due to a light wind. The pilot suddenly notices that the balloon must gain 20 m of altitude in order to clear the top of a hill 100 m to the east. Required:
a. How much time does the pilot have to make the altitude change without crashing into the hill?
b. What minimum, constant, upward acceleration is needed in order to clear the hill?
c. What is the horizontal component of the balloon’s velocity at the instant that it clears the top of the hill?
d. What is the vertical component of the balloon’s velocity at the instant that it clears the top of the hill?

a. t = 34.5 s

b. a = 0.034 m/s²

c. Vx = 2.9 m/s

d. Vfy = 1.16 m/s

Explanation:

a.

The pilot has the time, in which the balloon reaches the hill. So, we cab find it by applying the following formula, due to constant velocity of balloon:

S = Vt

t = S/V

where,

t = time to make altitude change = ?

S = Horizontal Distance between Balloon and Hill = 100 m

V = Horizontal Velocity of Balloon = 2.9 m/s

Therefore,

t = (100 m)/(2.9 m/s)

t = 34.5 s

b.

In order to calculate required upward acceleration, we use 2nd equation of motion:

S = Vit + (0.5)at²

where,

S = Vertical Distance = 20 m

Vi = Initial Vertical Velocity = 0 m/s (since, balloon moves horizontally in start)

t = time required = 34.5 s

a = minimum acceleration required = ?

Therefore,

20 m = (0 m/s)(34.5 s) + (0.5)(a)(34.5 s)²

a = (20 m/594.53 s²)

a = 0.034 m/s²

c.

Assuming the air friction to be negligible, the horizontal velocity remains constant. Therefore,

Vx = 2.9 m/s

d.

In order to find the vertical component of final velocity we use 1st equation of motion:

Vfy = Vfi + at

where,

Vfy = Vertical Component of Velocity at instant that it clears the top of hill = ?

Therefore,

Vfy = 0 m/s + (0.034 m/s²)(34.5 s)

Vfy = 1.16 m/s

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

i think the answer is a and b

i wish you luck