A hot-air balloon is rising upward with a constant speed of 2.53 m/s. When the balloon is 5.82 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
Vi = 2.53
Hi = 5.82
0 = Hi + Vi t - 4.9 t^2
4.9 t^2 - 2.53 t - 5.82 = 0
solve quadratic for t
eg : http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
To find the time it takes for the compass to hit the ground, we need to determine the time it takes for the balloonist to drop the compass and the time it takes for the compass to fall from the height of the balloon to the ground.
First, let's calculate the time it takes for the compass to fall from the height of the balloon to the ground.
We can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
Where:
h is the vertical distance traveled (5.82 m in this case),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken.
Rearranging the equation, we get:
t^2 = (2h) / g
Substituting the values, we have:
t^2 = (2 * 5.82 m) / 9.8 m/s^2
t^2 = 0.5946 s^2
Taking the square root of both sides, we find:
t = 0.771 s
So, it takes approximately 0.771 seconds for the compass to fall from the height of the balloon to the ground.
Now, let's consider the time it takes for the balloonist to drop the compass.
Since the speed at which the balloon is rising is constant at 2.53 m/s, the time it takes for the balloonist to drop the compass is equal to the time it takes for the compass to fall.
Therefore, the total time that elapses before the compass hits the ground is approximately 0.771 seconds.