Did you try googling it lol thats what i do if its a problem like that. sometimes there are websites that answer it you just have to look really hard
Answer:
When the person has swung all the way back (position A), the swing pauses a moment. ... As the swing continues forward, it gradually slows down. Its kinetic energy changes back to potential energy until it reaches the farthest point in its arc (position C).
Answer:
<em>The velocity of the camera just before it hit the ground = 35.97 m/s.</em>
Explanation:
Velocity: This can be defined as the ratio of the displacement of a body to the time. Velocity is a vector quantity, and as such it can be represented both in magnitude and direction.
From the equation of motion,
v² = u² + 2gs ................ Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.
Note: Before the velocity of the camera before it hits the ground = The final velocity of the camera.
<em>Given: u = 10.8 m/s, s = 60 m. g = 9.81 m/s.</em>
<em>Substituting into equation 1,</em>
<em>v² = 10.8² + 2(9.81)(60)</em>
<em>v² = 116.64+1177.2</em>
<em>v² = 1293.84 </em>
<em>v = √(1293.84)</em>
<em>v = 35.97 m/s.</em>
<em>Hence the velocity of the camera just before it hit the ground = 35.97 m/s.</em>
The force and the air resistance depends on the mechanical enserfy.
