A hot-air balloon has just lifted off and is rising at the constant rate of 1.51 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 13.00 m/s. If the passenger is 2.68 m above her friend when the camera is tossed, how high is she when the camera first reaches her?
To find out how high the passenger is when the camera first reaches her, we need to determine the time it takes for the camera to reach her.
First, let's calculate the time it takes for the camera to reach its maximum height. We can use the formula:
t = v / g
Where:
t is the time in seconds
v is the initial vertical velocity of the camera (13.00 m/s)
g is the acceleration due to gravity (-9.8 m/s^2)
Plugging in the values, we find:
t = 13.00 m/s / -9.8 m/s^2
Calculating this, we get:
t ≈ -1.33 s
Note: The negative sign indicates that the vertical velocity is going against gravity.
Now, to find the height of the camera when it reaches its maximum height, we can use the equation:
h = h₀ + v₀ * t + 0.5 * gt^2
Where:
h is the height
h₀ is the initial height (2.68 m)
v₀ is the initial vertical velocity (13.00 m/s)
t is the time (which we found to be -1.33 s)
g is the acceleration due to gravity (-9.8 m/s^2)
Plugging in the values, we get:
h = 2.68 m + 13.00 m/s * (-1.33 s) + 0.5 * (-9.8 m/s^2) * (-1.33 s)^2
Calculating this, we get:
h ≈ 6.28 m
Therefore, the height of the passenger when the camera first reaches her is approximately 6.28 m.