Hi there!
We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

There is NO initial vertical velocity, so:

Rearrange to solve for time:

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

He or she would enjoying doing a job as a caretaker
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
option C and D
Explanation:
When a stunt driver is rotating on the banked road the force which is responsible for the centripetal acceleration of the car is Frictional force and the Normal Force.
The Frictional force component of the banked road will protect the car from skidding.
And the normal force component will protect the car from toppling inward due to centripetal force acting on it.
Hence, the correct answer is option C and D
the Axis Perpendicular to the Length and Passing through the Center. Take a uniform thin rod AB whose mass is M and length is l. YY’ axis passes through the center of the rod and perpendicular to the length of the rod. We have to calculate the moment of inertia about this axis. YY’.
Suppose, small piece dx is at a distance x from the YY' axis. Hence, mass of length dx is =(
l
M
)dx The moment of inertia of small piece about the YY' axis
dl=[(
l
M
)dx]x
2
Therefore, the moment of inertia of the complete rod about the YY' axis.
b) Moment of Inertia about the axis perpendicular to the the length and passing through the Corner If I
CD
is the moment of inertia about the axis CD perpendicular to the length of a thin rod and passing through the point A then, by theorem of parallel axis;
l
CD
=l
YY
′
+Md
2
=
12
Ml
2
+M(
2
l
)
2
l
CD
=
3
Ml
2
.........(2)