Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f = = 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T =
T= =0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
All the answers are:
a) The time that will it take to rise to the highest point is 3.06 seconds.
b) The ball will rise to a height of 45.87 meters.
c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.
The time when the ball has 10 m/s downward is 1.02 seconds.
d) The displacement of the ball will be zero at 6.12 seconds.
e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.
f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.
g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
Let's calculate the values for each case.
a) At the highest point, the final velocity is 0, so we can use the following equation.
(1)
Where:
We know that v(i) = 30 m/s.
Solve it for t:
Hence, the time is 3.06 s.
b) At the highest point, the final velocity is 0, so we can use the following equation.
(2)
We know that the initial velocity is 30 m/s.
Solving it for h we have:
Then, the height is 45.87 m.
c) Using equation (1) we can find the time (t).
So, the time elapsed to get 10 m/s is:
We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.
This is the time when the ball has 10 m/s downward.
Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.
d) It will be when the ball returns to the ground.
The displacement will be zero after 6.12 s.
e) Here we need to find the time when v(f) is 15 m/s
The time when the v(f) is 15 m/s is 1.53 s.
f) Here, we need to find t when h = 45.87/2 m = 22.94 m
We can use the next equation:
[tex]h=v_{i}t-0.5gt^{2}/tex]
[tex]22.94=30t-0.5*9.81*t^{2}/tex]
Solving this quadratic equation, t will be:
[tex]t=0.90\: s/tex]
Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.
g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
Learn more about vertical motion here:
I hope it helps you!