A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that
at the time of the release the balloon was 98.8 m above the ground, determine the time, τ, it takes the bag to reach the ground from the moment of its release.
1 answer:
When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
You might be interested in
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force
Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = ( = g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.