A hot air balloon has just lifted off and is rising at the constant rate of 2.2 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 11.6 m/s. If the passenger is 2.5 m above her friend when the camera is tossed, how high is she when the camera reaches her on its way up?
Passenger height = 2.5m + vt
= 2.5m + (2.2m/s)t
Camera height = (11.6m/s)t + (1/2)(-g)t^2
The camera reaches the passenger when
Passenger height = Camera height
solve t for equal height, then plug t into either equation for the height.
You will find 2 answers for equal height. One is the time going up, and the second is if the time the camera would be at the passenger's height coming down (if the passenger missed the camera on its upward flight).
To solve this problem, we need to find the time it takes for the camera to reach the passenger on its way up. Using the equation of motion, we can find the time as follows:
1. First, let's find the time it takes for the camera to reach its maximum height. We can use the formula: v = u + at, where:
- v = final velocity (0 m/s at the highest point since the camera momentarily stops)
- u = initial velocity (11.6 m/s)
- a = acceleration (considering gravity, which is approximately -9.8 m/s^2)
- t = time
Rearranging the formula to solve for t, we have:
0 = 11.6 - 9.8t
9.8t = 11.6
t = 11.6 / 9.8
t ≈ 1.18 seconds
2. Next, let's find the height the camera reaches. We can use the formula: s = ut + 0.5at^2, where:
- s = displacement (unknown)
- u = initial velocity (11.6 m/s)
- a = acceleration (approximately -9.8 m/s^2)
- t = time (1.18 seconds)
Substituting the known values into the formula, we have:
s = (11.6 × 1.18) + (0.5 × -9.8 × 1.18^2)
s ≈ 6.83 meters
3. Now that we know the camera reaches a height of 6.83 meters, let's find the additional height the passenger has gained during that time. The passenger has been ascending in the hot air balloon at a rate of 2.2 m/s for 1.18 seconds. Using the formula: s = vt, where:
- s = displacement (unknown)
- v = velocity (2.2 m/s)
- t = time (1.18 seconds)
Substituting the known values into the formula, we have:
s = 2.2 × 1.18
s ≈ 2.59 meters
4. Finally, to determine the total height of the passenger when the camera reaches her, we need to add the additional height to the initial height of 2.5 meters. Adding the two values, we have:
Total height = 2.5 + 6.83 + 2.59
Total height ≈ 11.92 meters
Therefore, the passenger is approximately 11.92 meters high when the camera reaches her on its way up.