A hot air balloon is traveling vertically upward at a constant speed of 2.5 m/s. When
it is 30 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s^2
.
Answer in units of s
it is 30 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s^2
.
Answer in units of s
-
the up velocity of the package when released at a height of 30 m is 2.5 m/s.
the height reached by the package from this point is
h = v0^2/(2g)
(from setting potential energy = kinetic energy
1/2 mv^2 = mgh --> h = v0^2/(2g))
h = 2.5^2/(2*9.8) = 0.3189 m
the time needed to move these 0.3189 m is
t = v0/g = 2.5/9.8 = 0.2551 s
the maximum height of the package is therefore
hmax = 30 + 0.3189 m = 30.3189 m.
the time needed to hit the ground in a free fall from the max. height is found by
h = 1/2 gt^2
t = √(2h/g) = √(2*30.3189/9.8) = 2.487 s
---> the total time in the air is 0.2551 + 2.487 s = 2.7426 s <--- ans.
OG
the height reached by the package from this point is
h = v0^2/(2g)
(from setting potential energy = kinetic energy
1/2 mv^2 = mgh --> h = v0^2/(2g))
h = 2.5^2/(2*9.8) = 0.3189 m
the time needed to move these 0.3189 m is
t = v0/g = 2.5/9.8 = 0.2551 s
the maximum height of the package is therefore
hmax = 30 + 0.3189 m = 30.3189 m.
the time needed to hit the ground in a free fall from the max. height is found by
h = 1/2 gt^2
t = √(2h/g) = √(2*30.3189/9.8) = 2.487 s
---> the total time in the air is 0.2551 + 2.487 s = 2.7426 s <--- ans.
OG