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
Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + = m·g·h₂ +
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 +
5,061.96 J = 21.5 kg × v₂² + 53. kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.