A hotair balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sadndbag at the instant the balloon is 40.0 m above the ground. after it is released, the sandbag encounters no appreciable air drag. a. compute the position and velocity of the sandbag at .250 s and 1.00 s after its release. b. how many seconds after its release will the bag strike the ground? C. how fast is it moving as it strikes the ground? d. what is the greatest height above the ground that the sandbag reaches?
BOTO
To solve this problem, we can use the equations of motion for an object in free fall. The key is to recognize that once the sandbag is released, it will accelerate downward due to gravity, while the hot air balloon continues to rise vertically at a constant speed.
Let's solve each part of the problem step by step:
a) To compute the position and velocity of the sandbag at different times, we need to use the equations of motion. The equation for position (s) is:
s = ut + (1/2)at^2
where s is the position, u is the initial velocity, t is the time, and a is the acceleration.
Given:
Initial velocity, u = 0 m/s (because the sandbag is initially at rest)
Acceleration, a = 9.8 m/s^2 (due to gravity)
For t = 0.250 s:
s = (0)(0.250) + (1/2)(9.8)(0.250)^2
s = 0 + (1/2)(9.8)(0.0625)
s = 0 + 0.3075 m = 0.308 m
The position of the sandbag at 0.250 s is 0.308 m above the ground.
To find the velocity, we can use the equation:
v = u + at
v = 0 + (9.8)(0.250)
v = 2.45 m/s
The velocity of the sandbag at 0.250 s is 2.45 m/s downward.
Now, let's calculate the position and velocity of the sandbag at 1.00 s:
Using the equation for position:
s = (0)(1.00) + (1/2)(9.8)(1.00)^2
s = 0 + (1/2)(9.8)(1.00)
s = 0 + 4.9 m = 4.9 m
The position of the sandbag at 1.00 s is 4.9 m below the starting point (40.0 m above the ground).
Using the equation for velocity:
v = (0) + (9.8)(1.00)
v = 9.8 m/s
The velocity of the sandbag at 1.00 s is 9.8 m/s downward.
b) To find the time it takes for the sandbag to strike the ground, we need to determine when its position is equal to the ground level (0 m).
Using the equation for position:
0 = (0)(t) + (1/2)(9.8)(t^2)
0 = 0 + (4.9)(t^2)
Simplifying, we have:
t^2 = 0
This gives us t = 0s.
Therefore, the sandbag will strike the ground immediately upon release.
c) The speed of the sandbag just before hitting the ground can be calculated using the equation for velocity:
v = u + at
Given:
u = 0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = time of impact (from part b) = 0s
v = 0 + (9.8)(0)
v = 0 m/s
The sandbag has no velocity just before hitting the ground.
d) The greatest height above the ground that the sandbag reaches is the initial height of 40.0 m above the ground. Since the sandbag is in free fall, it will never attain a greater height.
So, the greatest height above the ground that the sandbag reaches is 40.0 m.
In summary:
a) At 0.250 s, the sandbag is at a height of 0.308 m above the ground and has a velocity of 2.45 m/s downward.
At 1.00 s, the sandbag is 4.9 m below the starting point and has a velocity of 9.8 m/s downward.
b) The sandbag will strike the ground immediately upon release (at t = 0s).
c) The sandbag has no velocity just before hitting the ground (v = 0 m/s).
d) The greatest height reached by the sandbag is the initial height of 40.0 m above the ground.