The velocity is given by:
V = √(Vx²+Vy²)
V = velocity, Vx = horizontal velocity, Vy = vertical velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for V:
V = √(6²+12²)
V = 13.42m/s
Now find the direction:
θ = tan⁻¹(Vy/Vx)
θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for θ:
θ = tan⁻¹(12/6)
θ = 63.4°
The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.
Answer:
c it is not accelerating on it's on but gravity pulls it there for velocity increases.
Answer:
7.09683 m
1.20285 s
2.4057 s
11.8 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)
From equation of motion we have
The maximum height above the ground that the ball reaches is 7.09683 m
Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s
The velocity just before it hits the ground is 11.8 m/s
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
