A hot air balloon is rising at a constant velocity of 13.3 m/s. When the balloon is at a height of 15 m, a heavy object is dropped overboard. IF air resistance on the object is neglected, how long does it take for it to reach the ground?
s = 15 + 13.3t - 4.9t²
solve for s=0
i don't get how to get rid of all the t's...
0=15+13.3t-4.9t^2
it is a quadratic equation, use the quadratic formula.
To find out how long it takes for the heavy object to reach the ground, we can use the equations of motion.
1. First, let's determine the time it takes for the object to reach the ground when dropped from a height of 15 m.
We know the initial velocity of the object is zero because it was initially stationary with respect to the balloon. The final displacement is the height from which it was dropped, which is 15 m. The acceleration acting on the object due to gravity is approximately 9.8 m/s^2.
Using the second equation of motion:
s = ut + (1/2)at^2
Where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.
We rearrange the equation:
15 = 0t + (1/2) * 9.8 * t^2
Simplifying the equation:
15 = 4.9t^2
Divide both sides by 4.9:
3 = t^2
Taking the square root of both sides:
t ≈ √3 ≈ 1.732 seconds
Therefore, it takes approximately 1.732 seconds for the heavy object to reach the ground when dropped from a height of 15 m.