Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.
Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8 
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion 
= 
- 2 a s
Put all the values in above formula we get,
⇒ 0 = 
- 2 × a ×0.2667
⇒ a = 1.2 
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
A. Acceleration that’s the answer I think so
Answer: Option A: 22.5 m
Explanation:
A stone is dropped from a tower. The initial velocity of the stone, u = 0.
Height of the tower, y = 100 m
Let the depth of the well be d.
Time taken of the drop of the stone, t = 5 s
The stone falls under acceleration due to gravity g = 9.81 m/s²
We will use second equation of motion:
s = u t + 0.5 g t²
⇒100 m + d = 0 + 0.5 × 9.81 m/s²× (5.00 s)²
⇒d = (122.6 - 100) m = 22.6 m ≅ 22.5 m
Thus, the correct option is A. The depth of the well is 22.5 m.
