Question

A hot-air balloon of diameter 12 m rises vertically at a constant speed of 11 m/s . A passenger accidentally drops his camera from the railing of the basket when it is 15 m above the ground.If the balloon continues to rise at the same speed, how high is the railing when the camera hits the ground?

Answer 1

To solve this problem we will apply the linear motion kinematic equations. First we will determine the time it takes for the camera to fall to the floor. Later with that time, we will calculate from the initial position when the balloon ascended.

$h = v_ot-\frac{1}{2}gt^2$

Here

$v_0$  = Initial velocity

g= Acceleration due to gravity

t = time

Replacing with our values

$-15=11t-\frac{1}{2}9.8t^2$

Solving for the time,

$t = 3.2s$

Now with the previous position we have that the balloon has ascended around to,

$h_r=h_0 +v_0t$

$h_r = 15+11(3.2)$

$h_r = 50.2m$

Answer 2

The height that  is the railing when the camera hits the ground should be considered as the 50.2 m.

Calculation of the height:

Since we know that

$h = v_ot - 1/2gt^2$

Here

$v_o$ = initial velocity

g= Acceleration due to gravity

t = time

Now

$-15 = 11t - 1/29.8t^2$

So, t = 3.2 seconds

So,

the height is

= 15 + 11(3.2)

= 50.2 m

Hence, The height that  is the railing when the camera hits the ground should be considered as the 50.2 m.

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