1. Sandbag is dropped from rest from a hot air balloon at an altitude of 124 m. A horizontal wind is blowing, and the wind gives the sandbag a constant horizontal acceleration of 1.10m/s2. (a) Show that the path of the sandbag is a straight line. (b) How log does it take to hit the ground? (c) With what speed does it hit the ground?
(a) Well, if the wind always pushes the sandbag in the same horizontal direction, it means the sandbag will continue to move in that direction without any deviation. So, the path of the sandbag will indeed be a straight line. Unless, of course, it has other ideas and decides to take a detour to visit its sandbag friends, but I highly doubt that's the case here.
(b) To find out how long it takes for the sandbag to hit the ground, we can use the formula:
time = square root of [2 * (vertical distance) / (acceleration due to gravity)],
which in this case is:
time = square root of [2 * 124m / 9.8m/s^2].
I'll leave you to calculate the exact value because I don't want to steal the thunder from your calculator.
(c) Now, to determine the speed at which the sandbag hits the ground, we can use the equation:
final velocity = initial velocity + (acceleration * time).
The initial velocity in this case is 0 m/s (as it was dropped from rest) and the acceleration is the acceleration due to gravity, which is 9.8 m/s^2. Now, just plug in the values and calculate the final velocity.
Voila! You've got your answer. Well, that was a bit technical, but don't worry, I'm always here to clown around and lighten the mood.
(a) To show that the path of the sandbag is a straight line, we need to demonstrate that its vertical motion is independent of its horizontal motion.
Assuming the positive y-direction is upward, the only force acting vertically on the sandbag is gravity, which gives it a constant downward acceleration of 9.8 m/s^2. This acceleration is independent of the horizontal motion and remains constant throughout.
Since the sandbag has a constant vertical acceleration and no other forces acting on it vertically, we can conclude that its vertical motion is free-fall under gravity. This means that the path of the sandbag is a straight line.
(b) To find the time it takes for the sandbag to hit the ground, we can use the equation of motion for the vertical direction:
y = y0 + v0y*t + (1/2)*a*t^2
Here, y is the final position (ground level), y0 is the initial position (124 m), v0y is the initial vertical velocity (0 m/s since the sandbag is dropped from rest), a is the vertical acceleration (-9.8 m/s^2), and t is the time we want to find.
Plugging in the values, the equation becomes:
0 = 124 + 0*t + (1/2)*(-9.8)*t^2
Rearranging the equation, we have:
4.9*t^2 = 124
Simplifying further, we get:
t^2 = 124/4.9
t^2 = 25.30612245
Taking the square root of both sides, we get:
t = √(25.30612245)
t ≈ 5.03 s
So, it takes approximately 5.03 seconds for the sandbag to hit the ground.
(c) To find the speed with which the sandbag hits the ground, we can use the equation of motion for the vertical direction:
v = v0 + a*t
Here, v is the final velocity (which is what we want to find), v0 is the initial vertical velocity (0 m/s), a is the vertical acceleration (-9.8 m/s^2), and t is the time (5.03 s).
Plugging in the values, the equation becomes:
v = 0 + (-9.8)*5.03
v ≈ -49.294 m/s
The negative sign indicates that the velocity is directed downward.
So, the sandbag hits the ground with a speed of approximately 49.294 m/s.
To answer these questions, we'll need to use the equations of motion for both the horizontal and vertical components of the sandbag's motion. Let's break it down step by step:
(a) To show that the path of the sandbag is a straight line, we need to demonstrate that the horizontal and vertical motions are independent of each other. Since the wind provides a constant horizontal acceleration, the horizontal motion of the sandbag is uniformly accelerated.
The equation of motion for horizontal motion is given by:
x = x0 + v0xt + (1/2)axt^2
where:
x is the horizontal position of the sandbag
x0 is the initial horizontal position (which we can take as zero)
v0x is the initial horizontal velocity (which is zero as the sandbag is dropped from rest)
ax is the horizontal acceleration provided by the wind (1.10 m/s^2)
t is the time
Since the initial horizontal velocity is zero, and the horizontal acceleration is constant, the horizontal equation simplifies to:
x = (1/2)axt^2
This equation shows that the horizontal motion is solely determined by the horizontal acceleration and time. It does not depend on the vertical motion.
Therefore, we can conclude that the path of the sandbag is a straight line.
(b) To find the time it takes for the sandbag to hit the ground, we need to consider the vertical motion. We know that the sandbag is dropped from a height of 124 m, and we need to find the time it takes for it to fall to the ground.
The equation of motion for vertical motion, neglecting air resistance, is given by:
y = y0 + v0yt + (1/2)ayt^2
where:
y is the vertical position of the sandbag
y0 is the initial vertical position (124 m)
v0y is the initial vertical velocity (which is zero as the sandbag is dropped from rest)
ay is the acceleration due to gravity (-9.8 m/s^2)
t is the time
Setting y to zero (since the ground is at y = 0) and y0 to 124 m, we can rewrite the equation as:
0 = 124 - (1/2)(9.8)t^2
Rearranging and simplifying:
t^2 = (2 * 124) / 9.8
t^2 = 25.1429
Taking the square root of both sides:
t = √25.1429
t = 5.0143 s
Therefore, it takes approximately 5.0143 seconds for the sandbag to hit the ground.
(c) To find the speed at which the sandbag hits the ground, we can consider the vertical motion again. We know that the initial vertical velocity is zero, and we have already determined the time taken to hit the ground as 5.0143 seconds.
The final velocity (v) is given by the equation:
v = v0y + ayt
Since the initial vertical velocity (v0y) is zero, and the acceleration due to gravity (ay) is -9.8 m/s^2, we can simplify the equation to:
v = -9.8 * 5.0143
Calculating the value:
v = -49.0714 m/s
Since velocity is a vector quantity and we only have the magnitude, the negative sign indicates that the velocity is directed downwards.
Therefore, the sandbag hits the ground with a speed of approximately 49.0714 m/s in the downward direction.
(a) x" = 1.1
y" = -9.8
y" = -8.9 x"
dy/dt = -8.9 dx/dt
dy/dx = -8.9
since that is constant, y(x) is a line
(b) 4.9t^2 = 124
(c) speed s^2 = (dx/dt)^2 + (dy/dt)^2