Answer :
Answer:
1. 7.82 s
2. -76.7 m/s
3. 4067 N
Explanation:
Given:
y₀ = 300 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t and v
1.
y = y₀ + v₀ t + ½ at²
0 = 300 + (0) t + ½ (-9.8) t²
t = 7.82 s
2.
v² = v₀² + 2a (y − y₀)
v² = (0)² + 2 (-9.8) (0 − 300)
v = -76.7 m/s
3.
Draw a free body diagram. There are two forces on the container. Weight pulling down and drag force pulling up.
If the container falls at constant velocity, its acceleration is 0. Apply Newton's second law:
∑F = ma
D − mg = 0
D = mg
Given m = 415 kg:
D = (415) (9.8)
D = 4067 N
Round as needed.
Answer:
- [tex]t=7.82s[/tex].
- [tex]v_{f}=76.7 (m/s)[/tex] towards the ground.
- [tex]f_{air}=4067N[/tex].
Explanation:
In cases where we ignore the air resistance the problem is reduced to a free-fall problem. We are going to consider gravity positive since it is pointing in the same direction as the container falls.
1) The time it takes to the container to fall to the ground can be computed with the equation:
[tex]\Delta y=v_{0}t+0.5gt^{2}[/tex],
where [tex]\Delta y[/tex] is the distance the container falls, [tex]v_{0}[/tex] is the initial velocity of the container, [tex]g[/tex] is the gravity and [tex]t[/tex] is the time it takes to the container to fall [tex]\Delta y[/tex].
Since the container was dropped with no initial velocity our equation reduces to
[tex]\Delta y=0.5gt^{2}[/tex],
so the time it takes to fall is
[tex]t=\sqrt{\frac{\Delta y}{0.5g}}[/tex],
[tex]t=\sqrt{\frac{300}{0.5(9.8)}}[/tex],
[tex]t=7.82s[/tex].
2)This is still a free-fall problem, so we are going to use another free-fall equation:
[tex](v_{f})^{2}=(v_{0})^{2}+2g\Delta y[/tex],
since the initial velocity is 0
[tex](v_{f})^{2}=+2g\Delta y[/tex],
[tex]v_{f}=\sqrt{2g\Delta y}[/tex],
[tex]v_{f}=\sqrt{2(9.8)(300)[/tex],
[tex]v_{f}=76.7 (m/s)[/tex] towards the ground.
3)Now, since we are considering air resistance our problem becomes a Newton's laws of motion problem, specifically, Newton's second law of motion problem.
Recalling Newton's second law
[tex]F_{net}=ma[/tex],
where [tex]F_{net}[/tex] is the sum of all applied forces to a body, m is the body's mass and a is its acceleration.
In this case the forces are the weigth and the air resistance [tex]f_{air}[/tex], so
[tex]mg-f_{air}=ma[/tex].
We want to know the force that allows the container to fall at a constant speed, that means with no acceleration, so
[tex]mg-f_{air}=0[/tex],
[tex]f_{air}=mg[/tex],
[tex]f_{air}=415(9.8)[/tex],
[tex]f_{air}=4067N[/tex].