Question

A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the helicopter is 60.0 m above the ground. (a) How long will the camera take to reach the ground?

Answer 1

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 $\frac{m}{s}$.

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 $\frac{m}{s}$).

$V_{f} ^{2}$=$V_{0} ^{2}$ - 2 * g * h

Where:

$V_{f}$: final speed at maximum height.

$V_{0}$: initial speed when it falls from the helicopter.

g: gravity taken at 9.8 $\frac{m}{s^{2} }$

h: height reached from 60 m when leaving the helicopter

as $V_{f}$=0

0=$(12,5\frac{m}{s}) ^{2}$ - 2 * $9,8\frac{m}{s^{2} }$ * h

clear h:

h=$(12,5 \frac{m}{s^{2} } )^{2}$ / (2 * $9,8 \frac{m}{s^{2} }$)

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

$V_{f}$=$V_{0}$ - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as $V_{f}$=0

0=12.5 $\frac{m}{s}$ - 9.8 $\frac{m}{s^{2} }$ * t

clearing t

t=12.5 $\frac{m}{s}$ / 9.8 $\frac{m}{s^{2} }$

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y=$v_{0}*t+(\frac{g*t^{2} }{2} )$

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

$v_{0}$: initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 $\frac{m}{s^{2} }$

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

$v_{0}$=0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y=$\frac{(g*t^{2})}{2}$

Y*2=(g*$t^{2}$)

$\frac{Y*2}{g}$=$t^{2}$

$\sqrt{\frac{Y*2}{g} }$=t

Step 3: I replace the values with the incognites and get "t".

t=$\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }$

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!