A hot-air balloon is rising upward with a constant speed of 4.00 m/s. When the balloon is 4.77 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?
distancefalling=vi*t-4.9t^2
solve for t. distance falling is given, initial velocity is given.
To find the time it takes for the compass to hit the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance (height above the ground) = 4.77 m
u = initial velocity = 0 m/s (since the compass is dropped)
a = acceleration = 9.8 m/s^2 (due to gravity)
t = time
Since the balloon is rising upward with a constant speed of 4.00 m/s, the velocity of the compass is the same as the balloon, i.e., 4.00 m/s in the upward direction.
Using the equation of motion, we can rewrite it as:
s = ut + (1/2)at^2
4.77 = (4)(t) + (1/2)(-9.8)(t^2)
Simplifying the equation:
4.77 = 4t - 4.9t^2
Rearranging the equation to form a quadratic equation:
4.9t^2 - 4t + 4.77 = 0
Now we can solve the quadratic equation:
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation:
a = 4.9
b = -4
c = 4.77
t = (-(-4) ± √((-4)^2 - 4(4.9)(4.77))) / (2(4.9))
Calculating the values within the square root:
((-4)^2 - 4(4.9)(4.77))
= (16 - 94.96)
= -78.96
Since the value within the square root is negative, the equation does not have a real solution. This means that the compass will not hit the ground.
Therefore, there is no time elapsed before the compass hits the ground because it doesn't hit the ground in this scenario.