A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 6.1 m/s . When the balloon is 28 m above the ground, she accidentally releases the bag.
What is the speed of the bag just before it reaches the ground?
I got Vf^2= Vo^2 +2ad
Vf^2= 0+ (2*9.8) *28 = 548.8
Vf= Squareroot 548.8 = 23.4
But that answer is not correct. Can you please help me find my mistake and the right answer? THANK YOU!!!!!
what happened to vo? it was 6.1m/s, not zero.
To solve this problem, you correctly used the formula Vf^2 = Vo^2 + 2ad, which relates the final velocity (Vf), initial velocity (Vo), acceleration (a), and displacement (d) of an object.
In this case, the initial velocity (Vo) of the lunch bag is 0 m/s because it was released from rest. The acceleration (a) is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. The displacement (d) is 28 m, which is the height of the balloon above the ground.
So, your equation becomes Vf^2 = 0^2 + 2 * 9.8 * 28 = 548.8.
And to find the final velocity (Vf), you correctly took the square root of 548.8, which is 23.4 m/s.
Therefore, your calculations are correct, and the speed of the bag just before it reaches the ground is indeed 23.4 m/s. If you received a different answer, there might have been an error in your calculations or data entry. I recommend checking your work and rechecking the equation to ensure accuracy.