A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.
We can solve this problem using the equations of motion.
Let's consider the motion of the sandbag in the vertical direction. Since it is released with a constant speed and no air drag, it will continue to rise vertically with a constant velocity.
We know the initial height of the sandbag is 40.0 m, and its upward velocity is 5.00 m/s. We want to find the position of the sandbag after 0.250 s.
The formula we can use is:
Position = Initial position + (Velocity × Time) + (0.5 × Acceleration × Time^2)
In this case, the initial position is 40.0 m, the velocity is 5.00 m/s, the acceleration is -9.8 m/s^2 (negative because it's in the opposite direction of the motion), and the time is 0.250 s.
Plugging in these values, we get:
Position = 40.0 m + (5.00 m/s × 0.250 s) + (0.5 × -9.8 m/s^2 × (0.250 s)^2)
Position = 40.0 m + 1.25 m + (-0.30625 m) = 41.94 m
Therefore, the position of the sandbag at 0.250 s after its release is 41.94 m.
To solve this problem, we need to calculate the height of the sandbag when 0.250 seconds have passed since its release. Since the balloon is rising vertically with a constant speed of 5.00 m/s, the sandbag will also be moving with this speed.
We can start by determining the initial velocity of the sandbag. The initial velocity of the sandbag is the same as the balloon's velocity, which is 5.00 m/s upwards.
Next, we can use the equation for the displacement of an object with constant velocity to find the height of the sandbag after 0.250 seconds. The equation is:
Δy = v₀t + (1/2)at²
Where:
Δy is the displacement (change in height)
v₀ is the initial velocity
t is the time
a is the acceleration (which is zero in this case since there is no appreciable air drag)
Plugging in the values:
v₀ = 5.00 m/s (upwards)
t = 0.250 s
Δy = (5.00 m/s)(0.250 s) + (1/2)(0 m/s²)(0.250 s)²
Δy = 1.25 m + 0 m
Δy = 1.25 m
Therefore, after 0.250 seconds, the sandbag will be 1.25 meters above its initial position. Since the initial position was 40.0 m above the ground, the position of the sandbag at 0.250 s after its release will be 40.0 m + 1.25 m = 41.25 m above the ground.