A hot air balloon is 2.5m above the ground rising at 2m/s. A camera is thrown from the ground at 13m/s. At what height do the camer and balloon meet.
Need to know the correct formulas and how to insert the data. Stuck on this question for homework.
To solve this problem, you can use the concept of relative motion.
Let's assume the initial height of the hot air balloon is denoted as H0 and the initial height of the camera is denoted as h0. We want to find the height at which the camera and the balloon meet.
The height of the balloon can be represented as H = H0 + (2m/s * t), where "t" represents the time in seconds. Similarly, the height of the camera can be represented as h = h0 + (13m/s * t).
The point at which the camera and the balloon meet is when H = h. Therefore, we can set up the following equation:
H0 + (2m/s * t) = h0 + (13m/s * t)
Now, solve for "t":
H0 - h0 = (13m/s - 2m/s) * t
t = (H0 - h0) / (13m/s - 2m/s)
Since we are given that the balloon is initially 2.5m above the ground, H0 = 2.5m. Additionally, the camera is launched from the ground, so h0 = 0m. Substituting these values into the equation, we have:
t = (2.5m - 0m) / (13m/s - 2m/s)
t = 2.5m / 11m/s
t ā 0.227 seconds
Now that we have the time, we can find the height at which the camera and balloon meet by substituting the value of "t" into either H or h equations. Let's use the equation for H:
H = H0 + (2m/s * t)
H = 2.5m + (2m/s * 0.227s)
H ā 2.955m
Therefore, the height at which the camera and balloon meet is approximately 2.955 meters above the ground.
Make sure to check the accuracy of the calculations and apply appropriate rounding if necessary.