When a sandbag is dropped from a balloon hovering 1.3 m above the ground, it hist the ground at 5.0 ms⁻¹. On another occasion, the sandbag is released from the balloon which is rising at 7.0 ms⁻¹ when 1.3 m above the ground. There is also a horizontal cross wind of 5.0 ms⁻¹.
At what speed does the sandbag hit the ground?
The speed at which the sandbag hits the ground is 12.0 ms⁻¹. The vertical velocity of the sandbag is 7.0 ms⁻¹ (the speed of the balloon) plus the 5.0 ms⁻¹ of the cross wind, for a total of 12.0 ms⁻¹.
To find the speed at which the sandbag hits the ground, we need to consider the vertical and horizontal components of its motion separately.
In the first scenario, when the sandbag is dropped from a stationary balloon, we only need to consider the vertical motion.
Given:
- Initial velocity in the vertical direction (u): 0 m/s (since the balloon is stationary)
- Final velocity in the vertical direction (v): -5.0 m/s (negative because it's downward motion)
- Displacement in the vertical direction (s): -1.3 m (negative because it's downward motion)
- Acceleration due to gravity (g): 9.8 m/s² (downward motion)
We can use the equation of motion to solve for the time it takes for the sandbag to hit the ground:
v = u + gt
Rearranging the equation:
t = (v - u) / g
Substituting the given values:
t = (-5.0 - 0) / 9.8
t ≈ -0.51 seconds
Since time cannot be negative, we take the absolute value of the time: t ≈ 0.51 seconds.
Now, in the second scenario, when the sandbag is released from a rising balloon with a crosswind, we need to consider both the vertical and horizontal motion.
Given:
- Initial velocity in the vertical direction (u): 7.0 m/s (upward motion of the balloon)
- Final velocity in the vertical direction (v): -5.0 m/s (downward motion of the sandbag)
- Displacement in the vertical direction (s): -1.3 m (negative because it's downward motion)
- Acceleration due to gravity (g): 9.8 m/s² (downward motion)
Again, we can use the equation of motion to solve for the time:
v = u + gt
t = (v - u) / g
t = (-5.0 - 7.0) / 9.8
t ≈ -1.63 seconds
Since time cannot be negative, we take the absolute value of the time: t ≈ 1.63 seconds.
Now, let's find the horizontal distance traveled by the sandbag due to the crosswind.
Given:
- Horizontal velocity (v_h): 5.0 m/s (crosswind velocity)
- Time (t): 1.63 seconds
We can use the equation:
s_h = v_h * t
Substituting the given values:
s_h = 5.0 * 1.63
s_h ≈ 8.15 meters
Now, we can calculate the resultant velocity at which the sandbag hits the ground by considering both the vertical and horizontal components of its motion.
Resultant velocity (V) = sqrt((v_h)^2 + (v_v)^2)
Substituting the values:
V = sqrt((5.0)^2 + (-5.0)^2)
V = sqrt(25 + 25)
V = sqrt(50)
V ≈ 7.07 m/s
Therefore, the sandbag hits the ground with a speed of approximately 7.07 m/s.
To find the speed at which the sandbag hits the ground, we need to combine the vertical velocity caused by the balloon's motion and the crosswind with the acceleration due to gravity.
Let's break down the problem into its components:
1. Vertical Motion:
When dropped from a balloon, the sandbag falls vertically under the influence of gravity. We can use the equations of motion to solve for the vertical velocity.
The equation for vertical velocity can be written as:
vf = vi + at
where,
vf = final vertical velocity (unknown)
vi = initial vertical velocity (given as 0 m/s as the sandbag is initially at rest)
a = acceleration due to gravity = 9.8 m/s^2
t = time taken to reach the ground (unknown)
From the problem, we know that the vertical displacement is 1.3 m. Using the equation of motion for vertical displacement:
d = vit + (1/2)at^2
where,
d = vertical displacement = 1.3 m
vi = initial vertical velocity = 0 m/s
a = acceleration due to gravity = 9.8 m/s^2
t = time taken to reach the ground (unknown)
Rearranging the equation gives:
1.3 = (1/2)(9.8)t^2
Solving this equation will give us the value of t, which is the time taken for the sandbag to hit the ground. Once we have the value of t, we can substitute it back into the equation vf = vi + at to find the final vertical velocity.
2. Horizontal Motion:
The sandbag is also subject to a horizontal crosswind with a velocity of 5.0 m/s. However, since this crosswind does not affect the vertical motion, we can ignore it in this part of the problem.
3. Combining the Vertical and Horizontal Motion:
To find the total speed at which the sandbag hits the ground, we can use the Pythagorean theorem to combine the vertical and horizontal velocities into a single resultant velocity.
The formula for the resultant velocity is:
v = sqrt(v_horizontal^2 + v_vertical^2)
where,
v = resultant velocity (unknown)
v_horizontal = horizontal velocity (5.0 m/s)
v_vertical = vertical velocity (calculated in step 1)
Now, follow the steps outlined above to find the final speed at which the sandbag hits the ground.