Question

. Dr. Martin has decided to take a hot air balloon ride. The balloon leaves the ground and accelerates upward with an acceleration of g=4. After some time, t, Dr. Martin leans over the balloon's basket to take a picture with her cell phone. However, it is windy, and her phone is knocked out of her hands. How long does it take for her phone to reach the ground? Give your answer in terms of t. Here's a big hint: when Dr. Martin drops her phone, remember that is travelling upward with her in the balloon. (Don't be afraid! I know it seems weird to do physics without numbers, but it is important to get used to this type of mathematical reasoning.)

In order to get full credit for this problem you must show the proper problem-solving procedures, which at minimum includes a sketch with the important points in your motion and your known and unknown variables.

Answer 1

consider the motion in upward direction before cell phone falls

V_o = initial velocity of balloon = 0 m/s

a = acceleration = 4 m/s²

t = time of travel

Y = vertical displacement

V_f = velocity of balloon after time "t"

velocity of balloon after time "t" is given as

V_f = V_o + at

V_f = 0 + 4 t

V_f = 4t

height of the balloon after time "t" is given as

H = V_o t + (0.5) a t²

H = 0 t + (0.5) (4) t²

H = 2 t²

Consider the motion of the cellphone after it leaves the hand

V_o = initial velocity = 4t

Y = vertical displacement to reach the ground = - H = - 2t²

a = acceleration = - 9.8

t' = time taken to come back to ground

Using the equation

Y = V_o t' + (0.5) a t'²

- 2t² = (4t) t' + (0.5) (-9.8) t'²

(4.9) t'² - (4t) t' - 2t² = 0

t' = 4t sqrt(16t² + 39.2 t²) /9.8

t' = 4t 7.43 t /9.8

t' = 1.17t