A hot air balloon is ascending at the rate of 13 m/s and is 75 m above the ground when a package is dropped over the side. How long does the package take to reach the ground? With what speed does it hit the ground?
Vf^2=Vi^2+ 2gd where g=-9.8 and d=-75
How long?
Vf=Vi+at solve for t (remember Vf will be negative)
The height of the package vs. time is
y = 75 + 13 t - 4.9 t^2
Set y=0 and solve for t. Use the quadratic equation and take the solution for which t>0.
For the speed when it hits the ground, use the time t from the first step and
V = 13 - 9.8 t
To find the time it takes for the package to reach the ground and the speed it hits the ground, we can use the equations of motion. Specifically, we can use the equation for vertically accelerated motion.
The equation for the displacement of an object undergoing vertically accelerated motion is given by:
s = ut + (1/2)at^2
where:
s = displacement (in this case, the initial height above the ground)
u = initial velocity (in this case, the velocity of the ascending hot air balloon, which is 13 m/s upwards)
a = acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/s^2)
t = time
We want to find the time it takes for the package to reach the ground, so we set s (displacement) to zero. This is because the displacement when the package reaches the ground is zero.
0 = ut + (1/2)at^2
Substituting the known values:
0 = 13t + (1/2)(9.8)t^2
Simplifying:
(1/2)(9.8)t^2 + 13t = 0
To solve this quadratic equation, we can factor out t:
t((1/2)(9.8)t + 13) = 0
This equation has two solutions: t = 0 (which is not applicable in this case), and (1/2)(9.8)t + 13 = 0.
Solving for t:
(1/2)(9.8)t + 13 = 0
(1/2)(9.8)t = -13
t = -26/9.8 ≈ -2.65 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the time it takes for the package to reach the ground is approximately 2.65 seconds.
To find the speed at which the package hits the ground, we can use the equation for final velocity in vertically accelerated motion:
v = u + at
Substituting the known values:
v = 13 + (9.8)(2.65)
v = 13 + 25.37
v ≈ 38.37 m/s
Therefore, the package hits the ground with a speed of approximately 38.37 m/s.