A rock falls from rest off a cliff and hits the ground in 2 seconds. If there is no air resistance, determine the rock's velocit
1 answer:
Answer:
19.6m/s
Explanation:
A Rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.
here in our case because object starts off from rest and
is acceleration because of gravity ( Motion under gravity).
and of course t = 2 second.
Now by substituting all this information in equation of motion we get.
that would be the velocity of rock as it would hit the ground.
Note! We have assumed that there is no air resistance.
A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.
here in our case because object starts off from rest and is acceleration because of gravity ( Motion under gravity).
and of course t = 2 seconds.
Now by substituting all this information in equation of motion we get.
V = 19.6m/s
that would be the velocity of rock as it would hit the ground.
Note! We have assumed that there is no air resistance.
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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:
Explanation:
A number of the form
can be re-written in the radical form as follows:
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
Then, we can rewrite 1250 as
So we can rewrite the expression as
Answer:
Approximately . (Assuming that the drag on this ball is negligible, and that
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.)
- Vertical: constant downward acceleration at
, starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: . Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of , accelerated downwards at
, and reached the ground after
.
Apply the SUVAT equation to find the vertical displacement of this ball.
.
In other words, the ball is below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be
.