1.velocity and acceleration
2.
3.inertia
4.
5.speed
Answer: 211.059 m
Explanation:
We have the following data:
The angle at which the ball leaves the bat
The initial velocity of the ball
The acceleration due gravity
We need to find how far (horizontally) the ball travels in the air: 
Firstly we need to know this velocity has two components:
<u>Horizontally:</u>
(1)
(2)
<u>Vertically:</u>
(3)
(4)
On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when
and the time
is half the time it takes the complete parabolic path.
So, if we use the following equation, we will find
:
(5)
Isolating
:
(6)
(7)
(8)
Now that we have the time it takes to the ball to travel half of is path, we can find the total time
it takes the complete parabolic path, which is twice
:
(9)
With this result in mind, we can finally calculate how far the ball travels in the air:
(10)
Substituting (2) and (9) in (10):
(11)
Finally:
Answer:
16.7 s
Explanation:
T= <u>Vf - Vo</u> a= <u>F</u>
a m
4,500 / 3000 = 1.5 (a)
30 - 5 / 1.5(a) = 16.7 s
Answer: C
Explanation:
Find the acceleration using this kinematic equation:

Now use this kinematic equation to find the displacement:

Answer:
5.31143691523 m/s²
Explanation:
m = Mass = 280 g
x = Displacement of spring = 21.7 cm
Time period

Angular velocity is given by


From Hooke's law

The acceleration due to gravity on the planet is 5.31143691523 m/s²
Yes, I have been able to satisfy my curiosity.