Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through
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(a) 296.6 m
The motion of the stone is the motion of a projectile, thrown with a horizontal speed of
and with an initial vertical velocity of
where we have put a negative sign to indicate that the direction is downward.
The vertical position of the stone at time t is given by
(1)
where
h is the initial height
g = -9.81 m/s^2 is the acceleration due to gravity
The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:
y(6.00 s) = 0
Substituting into eq.(1), we can solve to find the initial height of the stone, h:
(b) 176.6 m
The balloon is moving downward with a constant vertical speed of
So the vertical position of the balloon after a time t is
and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:
(c) 198.2 m
In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.
The horizontal speed of the rock is
So the horizontal distance travelled in t = 6.00 s is
Considering also that the vertical height of the balloon after t=6.00 s is
The distance between the balloon and the rock can be found by using Pythagorean theorem:
(di) 15.0 m/s, -58.8 m/s
For an observer at rest in the basket, the rock is moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer at rest in the basket is
Substituting time t=6.00 s, we find
(dii) 15.0 m/s, -78.8 m/s
For an observer at rest on the ground, the rock is still moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer on the ground is now given by
Substituting time t=6.00 s, we find