a hot air balloonist, rising vertically with a constant velocity of magnitude 5m/s releases a sandbag at an instant when the balloon is 40m above the ground. after it is released, the sandbag is in free fall.
a) compute position at .250s
b.)how many seconds after its release, will it strike the ground?
c.)is free fall related to inertia that much that i should consider the newtonian law when answering this question?
The height at time t after release is, in meters,
H = 40 + 5 t - (g/2)t^2
g = 9.8 m/s^2 in this case
(a) Plug t = 0.25 s into that formula
(b) Solve 40 + 5t - 4.9 t^2 = 0 to get the t when it his the ground.
(c) Yes, you must consider the Newtonian Laws of motion whether in free fall or not.
why is 5m/s used as initial velocity of sandbag? if inertia was to be considered, what would be the sandbags initial volume then?
i used the formula:
Yf-Yi=Vit + [(-gt^2)/2]
my Vi was 0m/s since i assumed that the sandbag was at rest.
When the sandbag is released, it has the 5 m/s upward velocity of the ballon that is carrying it.
The volume of the sandbag is not asked for, and has nothing to do with the problem. Perhaps you meant Velocity
How did the sandbag gain a velocity of 5m/s upward? I thought of the sandbag remained at rest?
i meant of velocity. let me reword that: if inertia was to be considered, what would be the sandbags initial velocity then?
thanks for enlightening me.
Inertia IS considered. That is why the sandbag retains an initial vertical velocity of 5 m/s when it is dumped overboard. Then it starts accelerating downward because of gravity. Before that, it was being pushed upward with a force equal to the weight, but whatever rope or shelf was holdng it to the ballon
To answer these questions, we need to consider the properties of motion, such as velocity, time, acceleration, and displacement. Let's break it down step by step:
a) To compute the position of the sandbag at 0.250 seconds, we need to determine how far it has fallen during this time.
First, we need to find the initial velocity of the sandbag when it was released. Since the balloon was rising vertically with a constant velocity of magnitude 5 m/s, the sandbag would also have this initial velocity.
The equation for displacement is given by:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
t = time
a = acceleration
Since the sandbag is in free fall, the acceleration is equal to the acceleration due to gravity (approximately 9.8 m/s^2), but acting downward.
Substituting the values into the equation, we have:
s = 5 m/s * 0.250 s + (1/2) * (-9.8 m/s^2) * (0.250 s)^2
Simplifying the equation:
s = 1.25 m - 0.30625 m
s ā 0.94375 m
Therefore, at 0.250 seconds after its release, the sandbag will have fallen approximately 0.944 meters.
b) To determine how many seconds after its release the sandbag will strike the ground, we can use the equation for displacement:
s = ut + (1/2)at^2
In this case, s represents the displacement of the bag when it reaches the ground. Since the ground is 40 meters below the release point (negative displacement), we can set s = -40 m, u = 5 m/s, and a = -9.8 m/s^2.
-40 = 5t + (1/2)(-9.8)t^2
Simplifying the equation and setting it equal to zero:
(1/2)(-9.8)t^2 + 5t - 40 = 0
This is a quadratic equation, which can be solved using various methods (factoring, quadratic formula, etc.). Solving this equation, we find that t ā 4.1153 seconds.
So, approximately 4.115 seconds after its release, the sandbag will strike the ground.
c) Yes, free fall is related to inertia, and understanding Newton's laws of motion is helpful to answer this question. Newton's first law states that an object at rest stays at rest, and an object in motion stays in motion with the same velocity unless acted upon by an external force. In free fall, the only external force acting on the sandbag is gravity. So, understanding the principles of Newton's laws helps us determine the motion of the sandbag in free fall.