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 19 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?

The railing is at 56.4 m above the ground when the camera reaches the ground.

Explanation:

Hi there!

Let's find how much time it takes the camera to reach the ground. The equation of the height of the camera is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The initial height of the camera is 19 m and we need to find at which time its height is zero. Since the camera is dropped while the balloon is rising, the initial velocity of the camera is the same as the velocity of the balloon:

h = h0 + v0 · t + 1/2 · g · t²

When the camera hits the ground, h = 0

0 = 19 m + 11 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 19 m + 11 m/s · t - 4.9 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 3.4 s (The other value is rejected because it is negative and time can't be negative).

Since the balloon rises at constant speed, the equation of height of the railing is as follows:

h = h0 + v · t

To find the height of the railing 3.4 s after it was at 19 m, we have to solve the equation with h0 = 19 m and t = 3.4 s:

h = 19 m + 11 m/s · 3.4 s

h = 56.4 m

The railing is at 56.4 m above the ground when the camera reaches the ground.