A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?
1 answer:
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the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:
- The range of motion is: A = 4.44 cm
The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.
x = A cos (wt + Ф)
w² = k/m
where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.
Let's find the angular velocity/
w=
w = 25.8 rad/s
Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:
v=
v= - A w sin (wt +Ф)
0 = -A w sin Ф
Ф= 0
They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time
Position.
4.00 = A cos 25.8t
Speed.
50.0 = - At 25.8 sin 25.8t
To solve the system, ;et's square and add.
Cos² 25.8t =
sin² 25.8t =
1 =
A =
A= 4.44 cm
In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:
The range of motion is: A = 444 cm
Learn more about oscillatory motion here: brainly.com/question/14311816
Explanation:
Given that,
Charge acting on the object,
Force acting on the object, (in downward direction)
(a) The electric force acting in the electric field is given by :
E is the electric field
E = 4.75 N/C
The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.
(b) The charge on the proton is,
The force acting on the proton is :
If the charge on the proton is positive, the force on the proton is in upward direction.
Hence, this is the required solution.