A=(vf-vi)/t d= Vi x t + at2/2 d=(vf2-vi2)/2a d=((Vi+vf)/2) t
1 answer:
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Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L=
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L =
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
Answer:
c) The slope is not constant and increases with increasing time.
Explanation:
The equation for the position of this particle (starting from rest is)
We can take derivative of this with respect to time t to get the equation of slope:
As time t increase, the slope would increases with time as well.