A ball is released from a hot air balloon moving downward with a velocity of
-10.0 meters/second and a height of 1,000 meters. How long did it take the ball to reach the surface of Earth? Given: g = -9.8 meters/second2.
h= v₀t + gt²/2
gt²+2v₀t-2•h = 0
9.8t²+2•10t- 2000=0
t=[20±sqrt{400+4•9.8•1000}]/2•9.8 =
=(20±199)/19.6 =>
t=11.2 s.
its 11.2s
To answer this question, we need to use the equation of motion that relates the initial velocity, time, and acceleration due to gravity. The equation is:
h = h0 + v0t + (1/2)gt^2
Where:
h - height at time t
h0 - initial height
v0 - initial velocity
g - acceleration due to gravity
t - time
We know the following information:
h = 0 (because the ball reaches the surface)
h0 = 1000 meters
v0 = -10 meters/second
g = -9.8 meters/second^2
Substituting these values into the equation, we get:
0 = 1000 + (-10)t + (1/2)(-9.8)t^2
Simplifying the equation, we have:
0 = 1000 - 10t - 4.9t^2
This is a quadratic equation. To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, the coefficients are: a = -4.9, b = -10, and c = 1000.
Substituting these values into the quadratic formula, we get:
t = (-(-10) ± √((-10)^2 - 4(-4.9)(1000))) / (2(-4.9))
Simplifying further:
t = (10 ± √(100 + 19600)) / (-9.8)
t = (10 ± √19700) / (-9.8)
Now, we can calculate the two possible values for t by substituting √19700 as approximately 140 into the equation:
t1 = (10 + 140) / (-9.8) ≈ 15.96 seconds
t2 = (10 - 140) / (-9.8) ≈ -14.08 seconds
Since time cannot be negative in this context, the ball takes approximately 15.96 seconds to reach the surface of the Earth.