Answer:
This is what I know, The angle of releases changes the relationship between the horizontal and vertical components of projectile. The idea angle of release is 45 degrees, assuming there is no air resistance and take off and landing points are the same height.
Explanation:
L = length of the meter stick = 1 m
h = height of center of mass of stick from bottom end on the floor = L/2 = 1/2 = 0.5 m
m = mass of the meter stick
I = moment of inertia of the meter stick about the bottom end
w = angular velocity as it hits the floor
moment of inertia of the meter stick about the bottom end is given as
I = m L²/3
using conservation of energy
rotational kinetic energy of meter stick as it hits the floor = potential energy when it is vertical
(0.5) I w² = m g h
(0.5) (m L²/3) w² = m g h
( L²) w² = 6g h
( 1²) w² = 6 (9.8) (0.5)
w = 5.4 rad/s
Your answer is a new experiment, since it's something that hasn't been tried before, or has, but resulted in with lots of errors.
Answer:
Explanation:
When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.
According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.
As we know that work is given as:
here:
force of gravity on the object (which is acting downwards)
displacement of the object
- Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.
