1) Data:
Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m
2) Questions: V at Y = 3m and X at Y = 3 m
3) Calculate components of the initial velocity
Vox = Vo * cos(37°) = 15.97 m/s
Voy = Vo * sin(37°) = 12.04 m/s
4) Formulas
Vx = constant = 15.97 m/s
X = Vx * t
Vy = Voy - g*t
Y = Yo + Voy * t - g (t^2) / 2
5) Calculate t when Y = 3m (first time)
Use g ≈ 9.8 m/s^2
3 = 12.04 * t - 4.9 t^2
=> 4.9 t^2 - 12.04t + 3 = 0
Use the quadratic equation to solve the equation
=> t = 0.28 s and t = 2.18s
First time => t = 0.28 s.
6) Calculate Vy when t = 0.28 s
Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s
7) Calculate V:
V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s
tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°
Answer: V ≈ 18.5 m/s, with angle ≈ 30°
8) Calculate X at t = 0.28s
X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m
Answer: X ≈ 4,5 m
Answer:
Explanation:
r(t) = A(cos wt i + sin wt f)
= A cos wt i + A sin wt j
x = A cos wt
y = A sin wt
radius r
r² = x² + y² ( This is equation of a circle with radius r )
= A² cos² wt +A² sin² wt
= A²
r = A
radius r = A
b )
speed = dr/dt
v = - Aw sinwt i + Aw coswt j
magnitude of velocity
I v I= Aw √(sin²wt + cos²wt)
= Aw ( constant )
acceleration
= dv / dt = - Aw² cos wt - Aw² sinwt
magnitude of acceleration
I a I = Aw²
= r w²
d ) centripetal force = m acceleration
m w² A
=
Answer:
0.8976 seconds
Explanation:
The period of oscillation for the simple harmonic motion can be found using the formula ...
T = 2π√(d/g)
where d is the displacement of the spring due to the attached weight, and g is the acceleration due to gravity.
__
For d = 0.20 meters, the period is ...
T = 2π√(0.20/9.8) ≈ 0.8976 . . . . seconds
_____
<em>Additional comment</em>
The formula for the oscillator period is usually seen as ...
T = 2π√(m/k)
where m is the mass in the system and k is the spring constant. The value of the spring constant is calculated from ...
k = mg/d
Using that in the formula, we find it simplifies to ...
Compute the ball's angular speed <em>v</em> :
<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s
Use this to find the magnitude of the radial acceleration <em>a</em> :
<em>a</em> = <em>v </em>²/<em>R</em>
where <em>R</em> is the radius of the circular path. We get
<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²
The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude
∑ <em>F</em> = <em>m</em> <em>a</em>
where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then
<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N
The best answer would be a "non-directional hypothesis"
