Question

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera from the railing of the basket when it is 18 mm 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. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

$h = v_0 t -\frac{1}{2} gt^2$

$-18 = 15*t + \frac{1}{2} 9.8*t^2$

$t = 3.98s$

Then the total distance traveled would be

$h = h_0 +v_0t$

$h = 18+15*3.98$

$h = 77.7m$

Therefore the railing will be at a height of 77.7m when it has touched the ground

Answer 2

The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

Calculation o fthe height:

Since we know that

$h = vt - 1/2gt^2\\\\-18 = 15*1 + 1/2*9.8*t^2$

t = 3.98s

Now the total distance should be

$= 18 + 15*3.98$

= 77.7 m

hence, The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

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