A hot-air balloon is ascending at the rate of 13 m/s and is 73 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?
v0=13 m/s
S=-73 m
g=-9.8 m/s²
Use
S=v0*t+(1/2)g*t²
to solve for t.
Initial kinetic energy, KE1
=(1/2)m*v0²
Initial potential energy, PE1
=mgh
where
h=73 m
m=mass of packet
Final kinetic energy, KE2
=(1/2)mV²
Equate KE1+PE1=KE2 and solve for V, the landing velocity.
To find the answers to these questions, we will use equations of motion. Specifically, we will use the kinematic equations of motion.
(a) To calculate the time it takes for the package to reach the ground, we can use the equation:
h = ut + (1/2)gt^2
where:
- h is the height of the package above the ground (73 m)
- u is the initial velocity of the package (which is zero)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken (which we need to find)
Rearranging the equation, we get:
(1/2)gt^2 = h
Substituting the values, we have:
(1/2)(9.8)t^2 = 73
Simplifying the equation, we get:
4.9t^2 = 73
Dividing both sides by 4.9, we get:
t^2 = 73/4.9
Taking the square root of both sides, we have:
t = √(73/4.9)
Evaluating the square root, we find:
t ≈ 3.033
Therefore, the package takes approximately 3.033 seconds to reach the ground.
(b) To find the velocity at which the package hits the ground, we can use the equation:
v = u + gt
where:
- v is the final velocity (which we need to find)
- u is the initial velocity of the package (which is zero)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken (which we already found)
Substituting the values, we have:
v = 0 + 9.8 * 3.033
Calculating the product, we get:
v ≈ 29.83 m/s
Therefore, the package hits the ground with a velocity of approximately 29.83 m/s.