A hot-air balloon is rising straight up with a speed of 5 m/s. A ballast bag is released from rest relative to the balloon when it is 8.2 m above the ground. How much time elapses before the ballast bag hits the ground?
I don't get what equations or what values to use for this equation. can anyone help me?
Intial height, initial velocity given, looking for time.
h=Ho +Vi*t-1/2 9.8 t^2
solve when h=0. You know Ho, Vi.
Notice it is a quadratic.
To solve this problem, we can use the equations of motion. In this case, since the hot-air balloon is rising vertically, we can treat it as one-dimensional motion.
First, let's define the given values:
Initial speed of the balloon, u = 5 m/s (upward)
Initial position of the ballast bag, s = 8.2 m (above the ground)
We need to find the time it takes for the ballast bag to hit the ground, t.
To do that, we can use the equation of motion:
s = ut + (1/2)gt^2
where:
s is the displacement or change in position,
u is the initial velocity,
t is the time,
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the displacement, s, is the initial position of the ballast bag relative to the ground, which is 8.2 m.
Thus, the equation becomes:
8.2 = 5t + (1/2)(9.8)t^2
Simplifying this equation, we get:
4.9t^2 + 5t - 8.2 = 0
This is a quadratic equation that can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 4.9, b = 5, and c = -8.2.
Substituting these values into the quadratic formula, we can find the two possible values for t. Since we're dealing with time, we discard negative values:
t = (-5 + √(5^2 - 4(4.9)(-8.2))) / (2(4.9))
t ≈ 1.84 seconds (approximately)
Therefore, it takes approximately 1.84 seconds for the ballast bag to hit the ground.